I love these solvers to check my work or to do tedious calculations. Any others I should know about? The best part is they keep getting better. In October I made a video A Problem WolframAlpha Can't Solve, But You Can (615 + x^2 = 2^y) : ruclips.net/video/DOISjFviqkM/видео.html. Amazingly a month later someone said that WolframAlpha now does give the correct answer--so perhaps my video alerted them to a bug, or they decided to update the solver. Either way, it's amazing the kind of math technology we have access to today, and the tools just keep getting better.
There is a big mistake : 0^0 HAVE A VALUE, it’s 1 ! 0^0 is only an undermined form only for LIMITS so while this question is not about limits we don’t need to check if y != 0 in case 2
As a former college math professor, I just wanted to weigh in on this channel. You youngsters who are spending time solving these are doing yourself an amazing favor. These are great and fun ways to teach yourself to think of creative and logical solutions that will pay you dividends your entire life. There aren’t too many things you can do now that you will look back on when you are in your 70’s and say “That was a good use of time” but these problems are one of them. The presentation is so good and they have a nice range of problems. I wish I had something this nice when I was a teenager. What a great gift MindYourDecisions has given you!
My thoughts exactly. It has been about 35 years since I calculated anything like this but I remember that working with math was fun at school. However in this video there is one jump I don't know how you got what you got and that is how you get (x-2)(x-5)=0 from x^2-7x+11=1? Can someone explain that to me? Sounds like that is something you can do just by looking at the math so it probably is simple thing that I have just forgotten, but I can't figure that out. So "No, I don't think I can" is quite real for me.
@@benhur2806 Thanks for the answer but that doesn't really give me solution to my problem. It's the factoring part that I'm missing. Any websites where I can refresh my math skills (pretty basic stuff I realize now, I have forgotten more than I was willing to admit). I feel like the time I first encountered Laplace transforms. It was in physics class, not math and that was the first time I or my classmates had seen the thing. "You put L in front of that and the whole equation mysteriously changed? What's going on?"
@@3characterhandlerequired look up the quadratic formula, it gives the root of any x^2 polynomial (numbers that if inserted into the polynomial make the equation equal to 0)
Vinícius de A Batista I have watched this video 3 Times And stil I don't* understand what and what is Point of this. I really want to understand Math but i have learning disability called f80.
I am 65 years old. I watch the videos on this channel at night when I can't sleep. I usually only get a partial solution but I got this one totally. I appreciate this channel. I feel it keeps me sharp for my accounting work and my woodworking side business
Another way to solve: (x^2-7x+11)^(x^2-13x+42)=1 log|x^2-7x+11|^(x^2-13x+42)=0 (x^2-13x+42)log|x^2-7x+11|=0 then,x^2-13x+42=0 or log|x^2-7x+11|=0 first equation: x^2-13x+42=0 (x-6)(x-7)=0 x=6, 7 second equation: log|x^2-7x+11|=0 x^2-7x+11=1 or -1 x=2, 5, 3, 4 So x=2, 3, 4, 5, 6, 7
A Q He took the exponential of both sides and ended up with the absolute value sign of the function equal to 1. And since the exponential function and natural logarithmic function are inverses of each other, they cancel each other and thus you can get rid of the natural log. Next, after he gets rid of the absolute value sign, he can consider 1 or -1. Also, just in case you’re wondering why the absolute value is there in the first place, it is because the logarithmic function is defined only for x > 0, that is, only positive numbers.
@Zoltan Stromberg the explanation is simple. There is a difference when you solve the equation in integers or in real numbers. For integers, all solutions are valid, because indeed, (-1)^2n=1. It is not the case for real numbers though, because when you write a^b, where both a and b are real, then you immediately assume a>=0. Indeed, if a
@Zoltan Stromberg but the problem is not in two possible values, it is much deeper. If we consider a power function a^b, we would like to operate with b, as we always do with numbers. For example, (-8)^(1/3)=(-8)^(2/6)=(-8)^(3/9), etc. However, there three values of (-8)^(1/3) = {-2; 1+i√3; 1-i√3} and 6 values for (-8)^(2/6) = {+/-2; +/-(1+i√3); +/-(1-i√3)}. Now you can see what happens. Even for rational powers we can generate as many solutions as we want. In complex analysis, this is generalized to a^b = exp(b*ln|a|*b*(2πin+i*arctan(theta)), where n takes all possible integers and yields to infinite number of solutions. Thus, we have only two regular approaches. 1. Postulate that we want to consider only real solutions, therefore a>=0 forever, and negative a are not ever considered. 2. Properly extend the expression to the complex plane and show all infinite solutions. In this video it was done something in between that is not rigorous and not right, because the (-1)^2k solution is from the different realm, and all other solutions are magically ignored.
wow, I aspire to learn all this stuff one day, someday haha. I can understand the ambiguity that you stated here, but I lack the knowledge to understand anything else as whatever I know about complex numbers for now, is that its "a hack for computing negative roots"- as my maths teacher explained it. I know that's not just all, and that it's a whole new field of maths, but that's just the educational system here, incompetent teachers who didn't achieve their desired life goals (and thus reluctantly became teachers) who don't care about the student's intellectual understanding. they just care that we can solve the textbook questions and use so-called "tricks" to solve MCQs. I'm 15 by the way. :)
@@udith exactly, who knows if there are non-trivial solutions Solving the base to equal 1 and the exponent to be 0 might not give the complete solution set!
I got 4 just from the thumbnail, but as soon as you mentioned there are 2 more solutions I paused to figure them out. Took a bit, I'll admit, but it was satisfying!
I almost got tricked by your sound saying 7 is the answer of multiplying all of the solutions. When you look carefully, the real answer should be 7! instead of 7 and the sentence missed a full stop
He forgot the full stop. So it means that he is yelling that the answer is 7!!!! It is important for him us to remember that the answer is SEVEN!!!! Not 7! - it is SEVEN!
@@RExN6900 No, it isn't! If you use ! as a factorial, you still need a full stop to finish the sentence. Well, you don't need - you even start your sentences without any capitals.
You are one of the best useful channels of RUclips in my opinion Thnx for giving these types of questions these questions fecinate me about know more things
I took the natural log of the equation and that allowed me to find 2,5,6,7 but it missed 3 and 4. I think it is because my method assumes that the argument of the natural log is positive.
x=3,4 are not missed if you take the natural log and don't restrict yourself to real numbers. (x^2 - 13x + 42) ln(x^2 - 7x + 11) = ln(1) = 0 + i2πk Let x^2 - 13x + 42 = 2k and ln(x^2 - 7x + 11) = iπ. This occurs when x^2 - 7x + 11 = -1 --> x=3,4.
I feel like I had a problem similar to this on a competition where we had an average of 4 minutes to solve each problem (but the questions were overall easier). I forgot about the -1 case, but I remembered it when I was doing the problem for this video!
Sounds like you did what I did: y^0 1^y And forgot about the third case. That's ok, old dogs can remember old tricks just as we can learn the new ones. 😃
I found 6,7 and 2,5 from cases 1,2 in 30 seconds and i thought i was right.... but i forgot case 3... I figured it out the moment you said that many computers only find 4 solutions.....
The task is to solve for all REAL numbers x, so exponentiation operation b^n is possible only if b > 0. That is why the correct answers are 2, 5, 6, 7 despite the fact that 3, 4 satisfies the equation, however, this operation is not defined. Correct me if I'm wrong.
The LHS expression doesn't have real values when x is in the range (2,5) except at x=3 and x=4. So the graph is discontinuous, the graph plotter cannot draw a smooth line for you to spot the points where the equation takes the target RHS value of 1, and a native real solver won't find those solutions. For an automated solver to find these solutions really requires it to work in the complex field, in which case you do get a nice continuous line that veers out of the real plane at (2,1), then cuts back through it at (3,1) and (4,1), then rejoins the real plane at (5,1).
@@GeorgeFoot The graph plotter really ought to let you know that part of the graph is outside the range of the function. That way, you could know to wonder if there are solutions in there, even if it couldn't find them.
@@iabervon The line vanishing without going off the top or bottom should be a clue, but there are ways to make it more obvious for sure. The real win would be if the plotter did realise that there were points in range at 3 and 4 and showed them visually.
yes... and no. If you define x as a real number, it implies that the power notation is just a shortcut for exp((x²-13x+42)ln(x²-7x+11)). Therefore x²-7x+11 can't be a negative number by definition.
When I checked the graph in Desmos, I found out that it included the solutions 3 and 4. The reason why you missed it was because the solutions were merely points at the axis.
I'm in year 10 and I was so close!. I got 7 and 6 by knowing that any number to the power of zero is equal to one, and using simplifying as a way to solve for x, but I never knew the two other methods. thanks for teaching me!
4:03 If is easy to see that the exponent term, x²-13x+42 is even for ANY integer number, as x² and 13x will inevitably always have the same parity (odd if x is odd and even if x is even). So there is not even a need to explicitly compute these values for the values x=2 and x=3.
I only had the two cases in the beginning and then you mentioned that there were six solutions and I immediately was like oh yeah there’s -1 as well🤦🏼♀️
Unfortunately you are wrong about 3 and 4 x^2-13x+42 equals 12 for x = 3, but (-1)^(9-39+42) is not the same as (-1)^12, cause to do this step you need this a^(b+c) = a^b*a^c, which for real numbers is correct only when a>0 Also for one problem you can't use two different definitions of same a^b and problem explicetly stated real number so there can not be any ambiguity.
He didnt use that formula a^(b+c), because these are parameters, but u can use numbers so (-1)^(9-39+42) equals to (-1)^12 The formula could be used like if x^(x^2-5x+6) = x^(x^2) * x^(6-5x)
@@НиколайПотехин-н1е no you can't, given the definition of functions and operators same as sqrt(a*b) != sqrt(a)*sqrt(b), you can't do that willy-nilly even if those are specific numbers
@@AndreyGuild well, not exaclty. The number you are doing sqrt at has to be >= 0, but when it comes to ^ function, it is not necessary, you can do like 5^(-9)
that caught me off guard i didn't even consider the other cases, I knew in a heartbeat about case 2, I was so up in my own ass I didn't even consider other possible answers
Yes, the period. All of his other sentences had proper punctuation (and "own!" at 1:44), so if that was supposed to be a factorial symbol (math needs unique symbols instead of recycling punctuation ¬_¬), then it would have been a hanging sentence without terminating punctuation.
One such question also came in JEE Main 2016, there they asked sum of all possible values of x and gladly only one sum was matching in the options. It was put under the category of complex numbers. I got it right in the 2nd attempt.😉 👍
Go to Desmos, graph the equation, replace x with another variable like n, create a slider for the variable, set slider step to 1, and slide it over all six solutions shown in the video.
Thanks for the puzzle, Presh! The formula will be 1 only when the base has absolute value 1 or the exponent is 0. Since the exponent factors into (x-6)(x-7), we have {6,7} as two solutions. For the base, we have to find roots of (x^2-7*x+11)=1 instead, which is (x^2-7*x+10), which is (x-5)(x-2), giving us two more solutions, {2,5}. Our remaining solutions must have abs(base) = 1 and be raised to an exponent that winds them around to 1. Since we only seek real values for x, our remaining solutions must have base=-1 and an even exponent. So we find roots of (x^2-7*x+11)=-1 which is (x^2-7*x+12) which is (x-3)(x-4), and both substitutions give us even exponents, which gives our remaining solutions {3,4} So in summary we have x in {2,3,4,5,6,7}. Cute.
Nah - These are all the solutions. To get 1 on the RHS, you would need (i^4k), where k is an integer. Which is simply (i^2)^2k - Or (-1)^2k. By including case 3, this has already been covered! :)
Here is something to consider, we do not want to consider cases involving complex numbers because we want real solutions. So for example, we do not want to solve x^2-7x+11=i because there are no real solutions.
There aren't anymore, but for all folks replying, it is possible for a quadratic with complex coefficients to have one real and one complex solution, so it should generally be checked.
Watching these mathemathical problems and their solutions are like watching Val Valentino/masked magician revealing many magic's secrets. We look for overly complicated solutions only to find out that the solutions are quite simple. Not to mention the misdirections on the problems.
No need. Just observe that a^0 = 1 and 1^a is 1, and for good measure you may take 0^0 as 1 because of the limiting behaviour. That should give enough info to solve this.
@@pritivarshney2128 why? You know the only combinations are 1^something, something^0 and (-1)^2. That gives you the 3 equations you need to solve, each of them having 2 solutions for x, so there you have all 6 of them.
Alternatively apply natural log on both sides and ln |x^2 - 7x + 11| has 2 + or - values From which you get the additional 2 roots setting x^2 - 7x + 11 = -1
@@themouseonthebike lmao, it said "for high school student in Massachusetts" not MIT lol, read carefully before judging, think of it if this is first time u see this kind of question, not after tons of practices.
The third case ((-1) ^ 2k = 1) First, convert the result of the base equation in negative: x² - 7x + 11= -1 and turn it into x² - 7x + 11 + 1 = 0 = x² - 7x + 12 = (x - 3)(x - 4) so the solutions are 3 and 4 and substitute the x of the exponent with the solutions to find out that its result is an even number, since no matter whether the number is positive or negative, if it's elevated to an even exponent, the result is always positive.
My experience: 1. Found the first two solutions (6,7) knowing that a^0=1 (if a is not 0) and solved. I was so proud and kept reproducing the video... Until it was said that there were 6 solutions... I was like :O Then I paused the video immediately and kept finding solutions. 2. I found out that 1^a=1 as well (I realized it lately I know, I should've thought it) then solved and got other two solutions (5,2) In my calculator there is an option for you to evaluate values on a function, so I entered the given function and evaluated it for integers between 2 and 7 (because my answers were 2, 5, 6, 7, and my calculator only could receive values in intervals)... I found out that my answers were correct (equal to 1) but then realized that, unexpectedly, for 3 and 4, the answer was 1 as well (remember 3, 4 were in the given interval)... ...So yeah, basically I spoiled myself the last two solutions... 3. After almost 5 minutes later I found out negative 1 can be counted as well but only if the power is even, then solved and checked and voilà, other two solutions (3,4). So... I can say that I am proud of myself... I guess
I found only 2,5,6,7 and didn't consider -1 to the power of an even number. I really like your videos and this one was amazing as always. Keep up the good work. Thanks for everything.
The desmos graph does get all the solutions, but you just can’t see it. For example, the graph wont show 3 even though it’s an answer. But that’s because it’s too small of a point to graph because number a little less than 3 like 2.9 or a little more like 3.1 is undefined. You can check that desmos does recognize the answer if you tell desmos to plug in 3 for x and then it will mark the point for you
It may also plot the graph of f(x) = x¹ᐟ³, taking it to be f(x) = ∛x, but it's up to you to decide whether it makes sense under your definition of rational powers. And the same goes for (-1)²·⁰ - if you treat it as a *real* power (that is exponentiation with real exponents), you need to check your definitions.
@@igortelnuk Did you know that x ^ 2 -13x + 42 when x belongs to real numbers, the result will also be a real number? And when the exponent is a real number, the base of the exponent in negative numbers is undefined. That is, there are simply no such numbers. (А вы в курсе что x^2 -13x+42 при x принадлежащем действительным числам, результат тоже будет действительное число? А когда показатель степени действительное число, основание степени в отрицательных числах не определено. То есть таких чисел просто нет.)
@@dmitriytoropov4918 Did I know? Of course, I did. But this is true only when we are talking about the concept of a function. Or a general definition of a degree with a real exponent. Tell me, please, does the expression (-1) ^ 12, or (-1) ^ 6, or (x ^ 2-7x + 11) ^ 12 make sense? Are the numbers "3" and "4" real? В курсе ли я? Безусловно, да. Но это верно лишь когда мы говорим о понятии функции. Или общем определении степени с действительным показателем. Вот, скажите, имеет ли смысл выражение (-1)^12, или (-1)^6, или (x^2-7x+11)^12? Являются ли действительными числа "3" и "4"?
@@igortelnuk Как только мы говорим о решении в действительных числах, то любые операции с ними тоже рассматриваются как операции с действительными числами. 3 и 4 - действительные числа, x^2-7x+11 - тоже будет действительным и равным -1. Но не положительные действительные числа нельзя возводить в степень, поэтому ответ на ваш вопрос "Вот, скажите, имеет ли смысл выражение (-1)^12, или (-1)^6" только один, НЕТ не имеет, т.к. в данном контексте -1 является действительным числом.
@@igortelnuk "But this is only true when we are talking about the concept of function. Or a general definition of a degree with a real exponent." I wonder where you got this heresy, give a link if it does not make it difficult)) General definition, because it works wherever there are no additional comments. That is, in your opinion, if we get in the equation (-1) ^ 12 where 12 is a real number, then everything including the general definition goes to hell? You can't even imagine the degree of your delusion))) Understand, in the case (-1) ^ 12 where 12 is a real number, you get a number that is not in mathematics, because there is no description (definition) for it. "Here, tell me, does the expression (-1) ^ 12, or (-1) ^ 6, or (x ^ 2-7x + 11) ^ 12 make sense?" Your question directly says that you do not understand the sets of numbers, because before the question you need to indicate on which set the problem is being considered (look in any mathematic book). "Are the numbers '3' and '4' valid?" Again misunderstanding)) of course they are real numbers, and not only natural and whole numbers. You propose to pick out integers on all graphs of a real variable. The graphs will then become intermittent)) "Но это верно лишь когда мы говорим о понятии функции. Или общем определении степени с действительным показателем." Интересно где вы набрались этой ереси, скинте ссылку если не затруднит)) Общее определение, на то и общее, потому что работает везде где нет дополнительных замечаний. То есть по вашему если мы получим в уравнении (-1)^12 где 12 действительное число, то все идет лесом в том числе и общее определение? Вы даже не представляете степень своего заблуждения))) Поймите, в случае (-1)^12 где 12 действительное число вы получаете число которого нет в математике, потому что для него нет описания(определения). "Вот, скажите, имеет ли смысл выражение (-1)^12, или (-1)^6, или (x^2-7x+11)^12?" Ваш вопрос прямо говорит что вы не разбираетесь в множествах чисел, потому что перед вопросом надо указать на каком множестве рассматривается задача(загляните в любой учебник). "Являются ли действительными числа "3" и "4"?" Опять непонимание)) конечно они являются действительными числами, а не только натуральными и целыми. Вы предлагаете на всех графиках действительного переменного выкалывать целые числа. Графики тогда станут прерывистыми))
In this very case, noted that (x^2-13x+42) is definitely an even number, so case 1 and case 3 should be the same idea. But mostly, many student would neglect case 3
This equation can EASILY be solved on ANY COMPUTER with Microsoft Excel or the Open Office Calc (similar to Excel). No, there is absolutely no need of any mathematics software besides that.
@@bhanuchhabra7634 There is no need of the existence of such a video for what I claimed to be real. Two columns are enough to solve this. i) Use the first to create a list of numbers starting from -N up to +N. Use a small increment, like 0.01. Start with numbers -N, then use for the second cell -N + 0.01. Use dragging to fill the entire column with numbers incrementing as I described. ii) In the next column top cell write the expression with X replaced by a reference to the first cell in the first column. Do not use cell anchors. Drag the cell bottom left handle to fill the entire column with results. iii) Those cells containing 1 as a result correspond to a solution.
@@foreverandalwaysgfriend6669 What? IIT JEE ADVANCED In India.... ....Asia.... ....On Earth... 😃 Sorry just kidding If u wanna know more Then you must Google it If u can understand English Bro
@@rbkstudios2923 I have heard about IIT JEE ADVANCED. and I have even seen a few question papers. To me, an outsider, it looks boring (because it heavily relies on memory). That is the hard part, that it relies on memory. Mathematics isn't about knowing a formula and being to use it in a limited amount of time. I feel that JEE does not represent the true beauty of a Math exam. Moreover I have read that many of the candidates prepare for the JEE by familiarising themselves with the problems from the many question papers they have collected. This itself supports the claim that the exam is more of a memory test. But I may be wrong if what I have read are false. But do compare with the Olympiad questions, those are the questions that help the students explore. --- But hey, if the exam works in finding the best minds in the country, who am I to judge it. --- To conclude: Many knows that the derivative of x^2 wrt x is 2x (just by plugging in n=2 in the generalised formula of the derivative for x^n). But how many can explain why it is 2x? ----
@@foreverandalwaysgfriend6669 No dude Although it's not false that it's preparation must include checking of previous years question papers As some typical questions are repeatedly asked in that exam with changed figures but the same method of solving Yet it has beautiful and logical question as well They ask application based questions So they expect us to remember basic formulation It is really something else I've realized that what we learn for entrance exam, in many countries it is taught in 2nd or 3rd year of B.Sc courses And as for the derivative I know that almost 90% of my batch mates could derive that with 1st Principles of Differentiation And more 65% could explain that with proper logic No offence bro I too am weak in mathematics And I sometimes also hate that people who practice previous years paper get almost 2 questions in seconds in the actual examination
I love these solvers to check my work or to do tedious calculations. Any others I should know about? The best part is they keep getting better. In October I made a video A Problem WolframAlpha Can't Solve, But You Can (615 + x^2 = 2^y)
:
ruclips.net/video/DOISjFviqkM/видео.html. Amazingly a month later someone said that WolframAlpha now does give the correct answer--so perhaps my video alerted them to a bug, or they decided to update the solver. Either way, it's amazing the kind of math technology we have access to today, and the tools just keep getting better.
Happy New year
There is a big mistake : 0^0 HAVE A VALUE, it’s 1 !
0^0 is only an undermined form only for LIMITS so while this question is not about limits we don’t need to check if y != 0 in case 2
Wow, luckily i have just learnt that kind of exam for a few days. So, easy peasy lemon squizzy :))
@@vinceguemat3751 Setting aside the validity of 0^0=1, there are no 0^0 scenarios involving integers.
My mathematics teacher had taught me the same, and almost in the same way also, this question is really amazing.
As a former college math professor, I just wanted to weigh in on this channel. You youngsters who are spending time solving these are doing yourself an amazing favor. These are great and fun ways to teach yourself to think of creative and logical solutions that will pay you dividends your entire life. There aren’t too many things you can do now that you will look back on when you are in your 70’s and say “That was a good use of time” but these problems are one of them.
The presentation is so good and they have a nice range of problems. I wish I had something this nice when I was a teenager. What a great gift MindYourDecisions has given you!
I gor 0.11 ^42
You are right, these vids are like intellectual breaks from regular life.
Thanks for weighing in Sir...
...When I'm 70 I think I'll be more glad for the moments I lived in than the math problems I solved. (I do love math though.)
Ok boomer
"You might not believe it, but the answer is 7!" what a nice way to put factorials and trick everyone lmao
How's that work?
@@barneystinson3358 yeah, i dont get it
@@rarelyseen2609 the solutions are 2 3 4 5 6 7, so its 7! (7 factorial), so just a little wordplay ;)
omg
I just spotted it
"A math problem most computer can't solve, but you can"
Me: No, I don't think I can
My thoughts exactly. It has been about 35 years since I calculated anything like this but I remember that working with math was fun at school. However in this video there is one jump I don't know how you got what you got and that is how you get (x-2)(x-5)=0 from x^2-7x+11=1? Can someone explain that to me? Sounds like that is something you can do just by looking at the math so it probably is simple thing that I have just forgotten, but I can't figure that out. So "No, I don't think I can" is quite real for me.
@@3characterhandlerequired Subtract 1 from both sides, for x^2-7x+10=0, which you can then factor as (x-2)(x-5)=0
@@benhur2806 Thanks for the answer but that doesn't really give me solution to my problem. It's the factoring part that I'm missing. Any websites where I can refresh my math skills (pretty basic stuff I realize now, I have forgotten more than I was willing to admit). I feel like the time I first encountered Laplace transforms. It was in physics class, not math and that was the first time I or my classmates had seen the thing. "You put L in front of that and the whole equation mysteriously changed? What's going on?"
@@3characterhandlerequired Try Khan Academy or MathTutorDVD
@@3characterhandlerequired look up the quadratic formula, it gives the root of any x^2 polynomial (numbers that if inserted into the polynomial make the equation equal to 0)
The 1st and 2nd solutions easy but damn, I completely passed over that 3rd case
I am the only person who has also considered i^(4*k) and (-i)^(4*k), but completely missed that raising to 0th power gives you 1.
I actually thought about that but for some reason I got 13 instead of 12 and got 5 and 8 as solutions instead of 3 and 4
same
@VolodyA! V Anarhist
"Solve for all *real* numbers x". ;p
@@deedeeen That didn't stop me. I cannot remeber why.
0:54 - "well, imagine you are stuck in this problem"
Ok, that part I got right
Vinícius de A Batista I have watched this video 3 Times And stil I don't* understand what and what is Point of this. I really want to understand Math but i have learning disability called f80.
*these math videos inspire and build confidence for people around the world*
I did this question in first 15 seconds..believe me I got the roots of these quadratic just by seeing them and (-1)^2k was obvious forme
This comment is one the best ones I have seen I a while
I really enjoyed this problem as an algebra teacher. The first solution that came to my mind was when the exponent was 0.
I am a student who last studied maths a few years ago and all I could think of was 1^y = 1 solution
I got x^0 or 1^x but couldnt think of -1
@@pimbe2511 I'm a doctor, and I got similar thoughts
Yeah and I was just thinking to solve x² - 13x + 42 = 0 and you got it
I am a math teacher's daughter and I found all :)))
“How can we solve this problem? Well imagine that you are stuck on this problem”
No need to imagine then
I am 65 years old. I watch the videos on this channel at night when I can't sleep. I usually only get a partial solution but I got this one totally. I appreciate this channel. I feel it keeps me sharp for my accounting work and my woodworking side business
It's amazing how simple and straightforward the solution is when you just take what is known and connect the dots!
Another way to solve:
(x^2-7x+11)^(x^2-13x+42)=1
log|x^2-7x+11|^(x^2-13x+42)=0
(x^2-13x+42)log|x^2-7x+11|=0
then,x^2-13x+42=0 or
log|x^2-7x+11|=0
first equation:
x^2-13x+42=0
(x-6)(x-7)=0
x=6, 7
second equation:
log|x^2-7x+11|=0
x^2-7x+11=1 or -1
x=2, 5, 3, 4
So x=2, 3, 4, 5, 6, 7
That's a very simple method Kyanos. Even I love to do that.
That's a very elegant way to solve this problem. Il like it.
Although the video’s solution was equally ingenious, your solution was much simpler and more intuitive. Thank you
Why you used -1 because log(-1) isn't 0
A Q He took the exponential of both sides and ended up with the absolute value sign of the function equal to 1. And since the exponential function and natural logarithmic function are inverses of each other, they cancel each other and thus you can get rid of the natural log. Next, after he gets rid of the absolute value sign, he can consider 1 or -1.
Also, just in case you’re wondering why the absolute value is there in the first place, it is because the logarithmic function is defined only for x > 0, that is, only positive numbers.
Case in point wolfram alpha is the best.
No, the human is the grand solver...
You bet it is!
There is one equation that wolfram alpha can't solve
There are many things wolfram alpha cant solve but it is the best solver available. Unless of course you kidnap a maths professor.
@@DarthStrider ....on second thoughts that kidnapped prof + Alpha would be best 😆
Not gonna forget about case 3 for the rest of my life
LOOL
Because it is a wrong solution :)
@Zoltan Stromberg the explanation is simple. There is a difference when you solve the equation in integers or in real numbers. For integers, all solutions are valid, because indeed, (-1)^2n=1. It is not the case for real numbers though, because when you write a^b, where both a and b are real, then you immediately assume a>=0. Indeed, if a
@Zoltan Stromberg but the problem is not in two possible values, it is much deeper. If we consider a power function a^b, we would like to operate with b, as we always do with numbers. For example, (-8)^(1/3)=(-8)^(2/6)=(-8)^(3/9), etc. However, there three values of (-8)^(1/3) = {-2; 1+i√3; 1-i√3}
and 6 values for
(-8)^(2/6) = {+/-2; +/-(1+i√3); +/-(1-i√3)}.
Now you can see what happens. Even for rational powers we can generate as many solutions as we want. In complex analysis, this is generalized to
a^b = exp(b*ln|a|*b*(2πin+i*arctan(theta)), where n takes all possible integers and yields to infinite number of solutions. Thus, we have only two regular approaches.
1. Postulate that we want to consider only real solutions, therefore a>=0 forever, and negative a are not ever considered.
2. Properly extend the expression to the complex plane and show all infinite solutions.
In this video it was done something in between that is not rigorous and not right, because the (-1)^2k solution is from the different realm, and all other solutions are magically ignored.
wow, I aspire to learn all this stuff one day, someday haha. I can understand the ambiguity that you stated here, but I lack the knowledge to understand anything else as whatever I know about complex numbers for now, is that its "a hack for computing negative roots"- as my maths teacher explained it. I know that's not just all, and that it's a whole new field of maths, but that's just the educational system here, incompetent teachers who didn't achieve their desired life goals (and thus reluctantly became teachers) who don't care about the student's intellectual understanding. they just care that we can solve the textbook questions and use so-called "tricks" to solve MCQs. I'm 15 by the way. :)
Of course I am one year late, but oh man, this is one example why math is gorgeous :D
My goodness I actually knew how to do this, that's a first
I also
I got only 4
@@udith exactly, who knows if there are non-trivial solutions
Solving the base to equal 1 and the exponent to be 0 might not give the complete solution set!
I sadly only equated the bases. But I tried.
I got the solutions correct that is 7 but I got 6 to be correct too
I got 4 just from the thumbnail, but as soon as you mentioned there are 2 more solutions I paused to figure them out. Took a bit, I'll admit, but it was satisfying!
This was amazing! I’ve learned the 3rd case (Yes!). The factorial joke at the end is top notch 👏
i agree this is indeed amazing factorial
Nice trick
We know that 2*3*4*5*6*7 = 7!
What a nice way
oh thanks
wowowwwooo…………………… thanks
there is variable too bruh
@@AhmadRizviYT lol no
U misunderstood the joke
Ohhhhh! 7 factorial!! 😅😁
Try using log in it! It will be wayyyyy easier that way!
That way, RHS will become 0 has log 1 is zero, then easy-peasy calculation
no, if you use log you don't get 6 solutions you only get 4 of them
You will get all 6 roots
Cos log value is the modules of any number which always has two + or - value
@Gonçalo Fernandes great
You can this without using log also
Log was my first thought too.
I got the 6 or 7 immediately but did not think of the other two possible solutions. Great stuff.
2,5,6,7 and spent 3 min.
And forgotten case3
You forgot the chance that bottom can be (-1) and top can be even number
Same here then when he said there is 6 solutions I paused it and took me anither 5 minutes to try the negative
3 and 4 work as well because the bottom is -1 and the top is even.
Me too. Only got the last 2 solutions after he insisted there were 6. A great problem, and good video.
me too
I almost got tricked by your sound saying 7 is the answer of multiplying all of the solutions. When you look carefully, the real answer should be 7! instead of 7 and the sentence missed a full stop
He forgot the full stop. So it means that he is yelling that the answer is 7!!!! It is important for him us to remember that the answer is SEVEN!!!! Not 7! - it is SEVEN!
That's literally his intention. He wanted to trick everyone by making people like you ( =dumb) think the ! is the full stop of the sentence
@@Rekko82 this '!' at the end of a number means the factorial of the number
@@RExN6900 He knows
@@RExN6900 No, it isn't! If you use ! as a factorial, you still need a full stop to finish the sentence. Well, you don't need - you even start your sentences without any capitals.
That was one of the coolest sulotions I've ever seen for an algebra question....thanks mindyourdesicion
You are one of the best useful channels of RUclips in my opinion
Thnx for giving these types of questions these questions fecinate me about know more things
I took the natural log of the equation and that allowed me to find 2,5,6,7 but it missed 3 and 4. I think it is because my method assumes that the argument of the natural log is positive.
yeah i did that too..
This is probably the method the program used, that's why they missed 2 solutions.
I did log too. Surprised it wasn't in the video.
Same here, but in my defense I did that mentally sitting on a bus to work.. Guess I could have figured it with a pen and paper
x=3,4 are not missed if you take the natural log and don't restrict yourself to real numbers.
(x^2 - 13x + 42) ln(x^2 - 7x + 11) = ln(1) = 0 + i2πk
Let x^2 - 13x + 42 = 2k and
ln(x^2 - 7x + 11) = iπ.
This occurs when x^2 - 7x + 11 = -1 --> x=3,4.
My main man, the TI-92, solved for all 6! Pretty good for a 26-year old graphing calculator.
Thank you Presh! I have a math exam today and I was able to solve this! I have a major confidence boost now😁
Ju
I feel like I had a problem similar to this on a competition where we had an average of 4 minutes to solve each problem (but the questions were overall easier). I forgot about the -1 case, but I remembered it when I was doing the problem for this video!
It took me 3 minutes to get 2, 5, 6, and 7 without pen and paper. I thought I nailed this one, but incomplete. 🙁
mission failed we'll get em next time
@@harrysvensson2610 Incomplete mission *≠* Failed mission
Sounds like you did what I did:
y^0
1^y
And forgot about the third case. That's ok, old dogs can remember old tricks just as we can learn the new ones. 😃
WiseOldDude, same.
I was like: set the exponent to 0 and base to 1. Then factor them and bam I‘m done.
Haven‘t thought of (-1)^2k thing though
Did exactly the same..
before watching:
x^2-13x+42=0
or
x^2-7x+11=1
resolts:
6 or 7 or 5 or 2
Here u go 🏆
I found 6,7 and 2,5 from cases 1,2 in 30 seconds and i thought i was right.... but i forgot case 3...
I figured it out the moment you said that many computers only find 4 solutions.....
Same
The task is to solve for all REAL numbers x, so exponentiation operation b^n is possible only if b > 0. That is why the correct answers are 2, 5, 6, 7 despite the fact that 3, 4 satisfies the equation, however, this operation is not defined. Correct me if I'm wrong.
Are you disputing that 3 and 4 are real numbers? You’re reading too far into it.
0:30
A similar question was given to students in JEE Mains 2016, with 2 minutes to solve !
yes bro
@Divyesh Raj ROFLMFAO 😆
well actually this is pretty easy, a^b=1, solve for a=1 and b=0 you'll get your answers, this is my guess before watching the video I got x=2,5,6,7
okay I actually forgot about exponent being even or odd
@@mihirshah1717 where are you from?
This one was actually pretty easy.
Edit: Damn it, I missed the third case!
Same, (-1)^2k is a weird way to say 1. But still valid nevertheless.
I missed the (-1)^y case but got it as soon as you said there were 6 solutions. Wolfram aplha now misses this one too!
INDIA CAT 2020, SLOT 1 | Had to solve under 2 minutes, thank god I am your regular viewer! Thanks always:)
Yeah I missed case 3. Really appreciate this sort of content!
I got a feeling that the only reason why this video was made was because of the joke in the end lol. Cracked me so well
I got 2,3,4,5,6,7 before watching.
EDIT: For once I did it right.
X2
David Nettey same
Good
@David Nettey same for me. I thought I did pretty well until I watched the rest of the video.
[(x-3)*(x-4)-1]^(x-6)(x-7)=1, so either make (x-3)*(x-4)-1 =1, or (x-6)(x-7)=0. therefore I got 2,5,6,7, but 3 and 4 are right too
Great video! I took all the three cases and answer matched.
But why didn't the graph show the last two solutions
It's because the last two solutions are in the interval where x^2-7x+11, which is inside log, becomes negative.
The LHS expression doesn't have real values when x is in the range (2,5) except at x=3 and x=4. So the graph is discontinuous, the graph plotter cannot draw a smooth line for you to spot the points where the equation takes the target RHS value of 1, and a native real solver won't find those solutions.
For an automated solver to find these solutions really requires it to work in the complex field, in which case you do get a nice continuous line that veers out of the real plane at (2,1), then cuts back through it at (3,1) and (4,1), then rejoins the real plane at (5,1).
@@GeorgeFoot The graph plotter really ought to let you know that part of the graph is outside the range of the function. That way, you could know to wonder if there are solutions in there, even if it couldn't find them.
@@iabervon The line vanishing without going off the top or bottom should be a clue, but there are ways to make it more obvious for sure. The real win would be if the plotter did realise that there were points in range at 3 and 4 and showed them visually.
@@GeorgeFoot Why do you know the graph plotter issue? I'm in math major but didn't know this. Also thank you for answering.
yes... and no. If you define x as a real number, it implies that the power notation is just a shortcut for exp((x²-13x+42)ln(x²-7x+11)).
Therefore x²-7x+11 can't be a negative number by definition.
Who says that ln(x^2 - 7x + 11) needs to be a real number? Take the complex logarithm
@@martinmach553 and so what ?
Huh?
When I checked the graph in Desmos, I found out that it included the solutions 3 and 4. The reason why you missed it was because the solutions were merely points at the axis.
I'm in year 10 and I was so close!. I got 7 and 6 by knowing that any number to the power of zero is equal to one, and using simplifying as a way to solve for x, but I never knew the two other methods. thanks for teaching me!
A similar question has been asked in JEE MAINS 2016.
Yea i just solved it
It was easy
Yas
Yes.... but that question has answer -3 and 10
Thats what
WolframAlpha to other Math tools: amateurs
bprp: believe in math, not in wolframalpha
Wolframalpha has left the chat
For the joke, I started typing 7! into my calculator. Then I noticed that was the joke
4:03 If is easy to see that the exponent term, x²-13x+42 is even for ANY integer number, as x² and 13x will inevitably always have the same parity (odd if x is odd and even if x is even). So there is not even a need to explicitly compute these values for the values x=2 and x=3.
I only had the two cases in the beginning and then you mentioned that there were six solutions and I immediately was like oh yeah there’s -1 as well🤦🏼♀️
I accidentally figured out case 3 when I messed up my algebra figuring out case 1.
"there are no accidents"
-master oogway
Task failed successfully!
Unfortunately you are wrong about 3 and 4
x^2-13x+42 equals 12 for x = 3, but (-1)^(9-39+42) is not the same as (-1)^12, cause to do this step you need this a^(b+c) = a^b*a^c, which for real numbers is correct only when a>0
Also for one problem you can't use two different definitions of same a^b and problem explicetly stated real number so there can not be any ambiguity.
He didnt use that formula a^(b+c), because these are parameters, but u can use numbers so (-1)^(9-39+42) equals to (-1)^12
The formula could be used like if x^(x^2-5x+6) = x^(x^2) * x^(6-5x)
@@НиколайПотехин-н1е no you can't, given the definition of functions and operators same as sqrt(a*b) != sqrt(a)*sqrt(b), you can't do that willy-nilly even if those are specific numbers
@@AndreyGuild well, not exaclty. The number you are doing sqrt at has to be >= 0, but when it comes to ^ function, it is not necessary, you can do like 5^(-9)
Even if you had to separate them it would work. (-1)^(9-39+42)=[(-1)^9][(-1)^-39][(-1)^42]=(-1)(1/-1)(1)=(-1)(-1)(1)=1
that caught me off guard i didn't even consider the other cases, I knew in a heartbeat about case 2, I was so up in my own ass I didn't even consider other possible answers
Well I got confused when I ignored "!" and found the product to be 5040😂😂
this symbol ! means to multiply all numbers under 7 which is 2,3,4,5,6,7
Me too!
Well, now UK how much is 7!
Checked it with my TI Nspire CX CAS, it found all the solutions too.
Technically, it should uave been "You might not believe it, but the answer is 7!."
It's the same... like what? The dot? Xd
Yes, the period. All of his other sentences had proper punctuation (and "own!" at 1:44), so if that was supposed to be a factorial symbol (math needs unique symbols instead of recycling punctuation ¬_¬), then it would have been a hanging sentence without terminating punctuation.
You are right, but as you know, he wanted to trick us, it was ok.🙂🙂🙂
THE ANSWER IS SEVEN!!!!!
Exactly, lots of riddles rely on improper punctuation as the key to the mystery.
Happy to see that the pure Python CAS, SymPy, reports all solutions with a straightforward call to `solve`.
One such question also came in JEE Main 2016, there they asked sum of all possible values of x and gladly only one sum was matching in the options. It was put under the category of complex numbers. I got it right in the 2nd attempt.😉 👍
Go to Desmos, graph the equation, replace x with another variable like n, create a slider for the variable, set slider step to 1, and slide it over all six solutions shown in the video.
Thanks for the puzzle, Presh!
The formula will be 1 only when the base has absolute value 1 or the exponent is 0. Since the exponent factors into (x-6)(x-7), we have {6,7} as two solutions. For the base, we have to find roots of (x^2-7*x+11)=1 instead, which is (x^2-7*x+10), which is (x-5)(x-2), giving us two more solutions, {2,5}. Our remaining solutions must have abs(base) = 1 and be raised to an exponent that winds them around to 1. Since we only seek real values for x, our remaining solutions must have base=-1 and an even exponent. So we find roots of (x^2-7*x+11)=-1 which is (x^2-7*x+12) which is (x-3)(x-4), and both substitutions give us even exponents, which gives our remaining solutions {3,4}
So in summary we have x in {2,3,4,5,6,7}. Cute.
Neat. I did think to consider the cases. Thank you.
This is actually easy question for those who are preparing for JEE exams.
Yeah, most of these are. It took my class group around 15 min to solve that Gaokao functions question.
Did I spell that correctly?
@@ashutoshmishra7429 jee mains is even very easy as compared to jee advanced.. 😂
They gave 30min per question. We solve such questions in 1min/question rate.
yup
That last question was really a good trick...
I thought it's 7 but actually it's 7!
Lol
It is not 7! . That was a joke🤣 actually 7! =5040.is it the answer?🤨
Aren't there more solutions if take into account the complex plane? It wasn't stated in the preconditions. i^2=-1 or am I making it too complex?
Nah - These are all the solutions.
To get 1 on the RHS, you would need (i^4k), where k is an integer. Which is simply (i^2)^2k - Or (-1)^2k. By including case 3, this has already been covered! :)
It said real roots 👍
Hilbert Barelds "Solve for all real numbers"
Here is something to consider, we do not want to consider cases involving complex numbers because we want real solutions. So for example, we do not want to solve
x^2-7x+11=i
because there are no real solutions.
There aren't anymore, but for all folks replying, it is possible for a quadratic with complex coefficients to have one real and one complex solution, so it should generally be checked.
At 3:54 why we are checking the exponent to be even. As we have considered the whole exponent to be -1.
This is the first time I answered a question on this channel without getting anything wrong or missing a possibility :D I'm improving
Watching these mathemathical problems and their solutions are like watching Val Valentino/masked magician revealing many magic's secrets. We look for overly complicated solutions only to find out that the solutions are quite simple. Not to mention the misdirections on the problems.
Solved this question in 1 minute.....
I m really happy for me😊😊
same!
I just solved the power x^2+13x+42=0. beacause anything power zero exept zero is one
THIS WAS GOOD, THANKS FOR MAKING THIS CHANNEL
I missed Case 3 and found 4 solutions...
Same
I found 10 solutions
Same I did:)
@@xxnotmuchxx how?
@@FoxSlyme oh nvm, i got 69 solutions
Me, an intellectual: take ln on both sides
EDIT: OMG THANKS FOR SO MANY LIKES! I HAVE NEVER GOTTEN THIS MANY BEFORE!
No need. Just observe that a^0 = 1 and 1^a is 1, and for good measure you may take 0^0 as 1 because of the limiting behaviour. That should give enough info to solve this.
@@dekippiesip Yes but for any other value except 1 on rhs we would have to take ln
@@pritivarshney2128 why? You know the only combinations are 1^something, something^0 and (-1)^2. That gives you the 3 equations you need to solve, each of them having 2 solutions for x, so there you have all 6 of them.
@@dekippiesip I am just joking that whenever you see variable in power you try to bring it down by taking ln
Me too, but I neglect the case 3...
At first this was kinda scary. But after like 2 secs I realized that the exponent should be equal to 0. Ridiculous.
The base being 1 leads to answers too
Alternatively apply natural log on both sides and ln |x^2 - 7x + 11| has 2 + or - values
From which you get the additional 2 roots setting x^2 - 7x + 11 = -1
Ohh, I have found only 4 solutions, the final case is very interesting
7!
It's not 7!🤨.........if it was 7! then answer would be 5040.don't you think😏
@PrakharRai I don't know if you're joking or not
Ммм, вершки и корешки
@@АртемКаспаров-ъ3и "наши" повсюду
This question is damn easy we have solved this type of question many a time in jee test
@Where is Jerry? the best university is oxford, cambridge or Harvard...
@@themouseonthebike lmao, it said "for high school student in Massachusetts" not MIT lol, read carefully before judging, think of it if this is first time u see this kind of question, not after tons of practices.
@@marcsonic01
? Not for math. And even in general, Princeton is a better school than all that you listed.
It's easy BECAUSE you have solved similar problems many times....
@@themouseonthebike Try coffin problems. They are truly brutal tho I solved some of them.
I got the solutions 2,5,6,7, didn't thought about solutions 3,4
The third case ((-1) ^ 2k = 1) First, convert the result of the base equation in negative: x² - 7x + 11= -1 and turn it into x² - 7x + 11 + 1 = 0 = x² - 7x + 12 = (x - 3)(x - 4) so the solutions are 3 and 4 and substitute the x of the exponent with the solutions to find out that its result is an even number, since no matter whether the number is positive or negative, if it's elevated to an even exponent, the result is always positive.
Use log e(ln) on both sides to get two cases, each having 2 real answers, first condition gives 6,7 and the second gives 2,5 and 3,4.
For x in (2,5), you have (negative real)^(real), so of course all the graphing software derps out.
what? negative real?
"So, how can we solve this problem? Well, imagine that you're stuck on this problem..."
Check!
My experience:
1. Found the first two solutions (6,7) knowing that a^0=1 (if a is not 0) and solved. I was so proud and kept reproducing the video... Until it was said that there were 6 solutions...
I was like :O
Then I paused the video immediately and kept finding solutions.
2. I found out that 1^a=1 as well (I realized it lately I know, I should've thought it) then solved and got other two solutions (5,2)
In my calculator there is an option for you to evaluate values on a function, so I entered the given function and evaluated it for integers between 2 and 7 (because my answers were 2, 5, 6, 7, and my calculator only could receive values in intervals)...
I found out that my answers were correct (equal to 1) but then realized that, unexpectedly, for 3 and 4, the answer was 1 as well (remember 3, 4 were in the given interval)...
...So yeah, basically I spoiled myself the last two solutions...
3. After almost 5 minutes later I found out negative 1 can be counted as well but only if the power is even, then solved and checked and voilà, other two solutions (3,4).
So... I can say that I am proud of myself... I guess
Artur Santi, I mean that is kinda like hard solving for all the solutions lol. Many can do that
Thanks for wasting my time!!
I found only 2,5,6,7 and didn't consider -1 to the power of an even number. I really like your videos and this one was amazing as always. Keep up the good work. Thanks for everything.
How about i^4n = 1 where n is an integer?
@@mesory the question only asks to find all the real solutions.
7! ... You are such a dork. I love it.
*Good to know that Americans are improving in STEM fields!*
we could also covert exponent to logarithmic and then log 1 to any base is zero, so we could equate x^2 -13x +42 with 0
In vietnam , u may solve this problem in 1'
1 minute?? And I thought the syllabus where I live is hard enough...
yeah. 1 minute.
How was that
1minute REALLY??????
This is not worth more than 30 seconds
In India we can solve it with only 30s
That was a damn easy question!!!
Damn I'd love to know where you got that pfp, beautiful 😍
I used logarithm on both sides of the equation and get x = 2,5,6,7 in 20 seconds but missed the negative case...
i did the same
You must use an absolute value to find all solutions
The desmos graph does get all the solutions, but you just can’t see it. For example, the graph wont show 3 even though it’s an answer. But that’s because it’s too small of a point to graph because number a little less than 3 like 2.9 or a little more like 3.1 is undefined. You can check that desmos does recognize the answer if you tell desmos to plug in 3 for x and then it will mark the point for you
It may also plot the graph of f(x) = x¹ᐟ³, taking it to be f(x) = ∛x, but it's up to you to decide whether it makes sense under your definition of rational powers. And the same goes for (-1)²·⁰ - if you treat it as a *real* power (that is exponentiation with real exponents), you need to check your definitions.
It's 7! (FACTORIAL) not number 7
No. 7! Means 7x6x5x4x3x2x1
@@aztecwarrior9729 7 factorial is the same as 7x6x5x4x3x2x1. Learn math boy
@@maulanax i think you must say this to Rohit Kanojia
r/woosh
because of function is not defined on interval X = [~2.5, ~4.5], values 3 and 4 are not solving the equation
Amazing! Only one man noticed that the base is greater equal to 0, when exponent is real number! Otherwise it is not defined. Congratulations!
Why no one has noticed the mistake? We solve this equation in real numbers! Right ansver is 2, 5, 6, 7.
Because, we solving it for all real numbers x, but not in real numbers for all equation
(Мы ищем действительные значения х, а не частей равенства)
@@igortelnuk Did you know that x ^ 2 -13x + 42 when x belongs to real numbers, the result will also be a real number? And when the exponent is a real number, the base of the exponent in negative numbers is undefined. That is, there are simply no such numbers.
(А вы в курсе что x^2 -13x+42 при x принадлежащем действительным числам, результат тоже будет действительное число? А когда показатель степени действительное число, основание степени в отрицательных числах не определено. То есть таких чисел просто нет.)
@@dmitriytoropov4918 Did I know? Of course, I did.
But this is true only when we are talking about the concept of a function. Or a general definition of a degree with a real exponent. Tell me, please, does the expression (-1) ^ 12, or (-1) ^ 6, or (x ^ 2-7x + 11) ^ 12 make sense? Are the numbers "3" and "4" real?
В курсе ли я? Безусловно, да.
Но это верно лишь когда мы говорим о понятии функции. Или общем определении степени с действительным показателем. Вот, скажите, имеет ли смысл выражение (-1)^12, или (-1)^6, или (x^2-7x+11)^12? Являются ли действительными числа "3" и "4"?
@@igortelnuk Как только мы говорим о решении в действительных числах, то любые операции с ними тоже рассматриваются как операции с действительными числами. 3 и 4 - действительные числа, x^2-7x+11 - тоже будет действительным и равным -1. Но не положительные действительные числа нельзя возводить в степень, поэтому ответ на ваш вопрос "Вот, скажите, имеет ли смысл выражение (-1)^12, или (-1)^6" только один, НЕТ не имеет, т.к. в данном контексте -1 является действительным числом.
@@igortelnuk "But this is only true when we are talking about the concept of function. Or a general definition of a degree with a real exponent." I wonder where you got this heresy, give a link if it does not make it difficult)) General definition, because it works wherever there are no additional comments. That is, in your opinion, if we get in the equation (-1) ^ 12 where 12 is a real number, then everything including the general definition goes to hell? You can't even imagine the degree of your delusion))) Understand, in the case (-1) ^ 12 where 12 is a real number, you get a number that is not in mathematics, because there is no description (definition) for it. "Here, tell me, does the expression (-1) ^ 12, or (-1) ^ 6, or (x ^ 2-7x + 11) ^ 12 make sense?" Your question directly says that you do not understand the sets of numbers, because before the question you need to indicate on which set the problem is being considered (look in any mathematic book). "Are the numbers '3' and '4' valid?" Again misunderstanding)) of course they are real numbers, and not only natural and whole numbers. You propose to pick out integers on all graphs of a real variable. The graphs will then become intermittent))
"Но это верно лишь когда мы говорим о понятии функции. Или общем определении степени с действительным показателем." Интересно где вы набрались этой ереси, скинте ссылку если не затруднит)) Общее определение, на то и общее, потому что работает везде где нет дополнительных замечаний. То есть по вашему если мы получим в уравнении (-1)^12 где 12 действительное число, то все идет лесом в том числе и общее определение? Вы даже не представляете степень своего заблуждения))) Поймите, в случае (-1)^12 где 12 действительное число вы получаете число которого нет в математике, потому что для него нет описания(определения). "Вот, скажите, имеет ли смысл выражение (-1)^12, или (-1)^6, или (x^2-7x+11)^12?" Ваш вопрос прямо говорит что вы не разбираетесь в множествах чисел, потому что перед вопросом надо указать на каком множестве рассматривается задача(загляните в любой учебник). "Являются ли действительными числа "3" и "4"?" Опять непонимание)) конечно они являются действительными числами, а не только натуральными и целыми. Вы предлагаете на всех графиках действительного переменного выкалывать целые числа. Графики тогда станут прерывистыми))
In this very case, noted that (x^2-13x+42) is definitely an even number, so case 1 and case 3 should be the same idea. But mostly, many student would neglect case 3
What ????
Very similar problem if not the same was asked in JEE mains exam in India only 3 minutes was given to solve!!!
I do remember doing that question. It was asked in JEE mains.
2 minutes...
r/iamverysmart
Matlab and Maple: *Are we a joke to you*
Php solvers have left the chat
Well Matlab gives you 2,5,6,7 if you force real solutions (otherwise it just jumps to the numeric solver and only finds 2)
What is Whis doing in Maths class? You should be fighting Merno.
Aren't you supposed to train goku?
@@brightjovanny that's just some bulma stuff
@@brightjovanny what's merno doesn't exist in my world?
The easiest questions on your channel😅.. I solved it in mind calculation
same dude. i guess we are both indian haha.
i didn't get the 3rd case though, that was clever!
@@TheMartian11 same bro
I was so proud of getting the 4 solutions by myself but i didnt think about the case 3. Nice one man..
And I just got 2 and 5 as answers by taking log at both side😂
You can't take log base 0 brooo. It ruins the equality!!
@@MarieAnne. place the 3 and 4 to the equation and check
This equation can EASILY be solved on ANY COMPUTER with Microsoft Excel or the Open Office Calc (similar to Excel). No, there is absolutely no need of any mathematics software besides that.
If you can point to any video, tutorial where excel can solve this
@@bhanuchhabra7634 There is no need of the existence of such a video for what I claimed to be real. Two columns are enough to solve this.
i) Use the first to create a list of numbers starting from -N up to +N. Use a small increment, like 0.01. Start with numbers -N, then use for the second cell -N + 0.01. Use dragging to fill the entire column with numbers incrementing as I described.
ii) In the next column top cell write the expression with X replaced by a reference to the first cell in the first column. Do not use cell anchors.
Drag the cell bottom left handle to fill the entire column with results.
iii) Those cells containing 1 as a result correspond to a solution.
It's very easy for matlab users
Heard About IIT JEE ADVANCED
This problem was nothing compared to the problems asked in that exam
?
@@foreverandalwaysgfriend6669 What?
IIT JEE ADVANCED
In India....
....Asia....
....On Earth...
😃
Sorry just kidding
If u wanna know more
Then you must Google it
If u can understand English Bro
@@rbkstudios2923 I have heard about IIT JEE ADVANCED. and I have even seen a few question papers. To me, an outsider, it looks boring (because it heavily relies on memory). That is the hard part, that it relies on memory. Mathematics isn't about knowing a formula and being to use it in a limited amount of time. I feel that JEE does not represent the true beauty of a Math exam.
Moreover I have read that many of the candidates prepare for the JEE by familiarising themselves with the problems from the many question papers they have collected. This itself supports the claim that the exam is more of a memory test.
But I may be wrong if what I have read are false.
But do compare with the Olympiad questions, those are the questions that help the students explore.
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But hey, if the exam works in finding the best minds in the country, who am I to judge it.
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To conclude: Many knows that the derivative of x^2 wrt x is 2x (just by plugging in n=2 in the generalised formula of the derivative for x^n). But how many can explain why it is 2x?
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@@foreverandalwaysgfriend6669
No dude
Although it's not false that it's preparation must include checking of previous years question papers
As some typical questions are repeatedly asked in that exam with changed figures but the same method of solving
Yet it has beautiful and logical question as well
They ask application based questions
So they expect us to remember basic formulation
It is really something else
I've realized that what we learn for entrance exam, in many countries it is taught in 2nd or 3rd year of B.Sc courses
And as for the derivative
I know that almost 90% of my batch mates could derive that with 1st Principles of Differentiation
And more 65% could explain that with proper logic
No offence bro
I too am weak in mathematics
And I sometimes also hate that people who practice previous years paper get almost 2 questions in seconds in the actual examination
@@foreverandalwaysgfriend6669
Where are you from?
'Coz it's night here with clock hands pointing at 2:16
I am not specialised in maths, but I love it. And these videos make me love maths even more
This question is very very important for CMI, ISI in India... I have solved it in class 9
Yes, and it was in the syllabus so