anyone else noticed that T is the golden ratio because R(n+2)=R(n+1)+R(n) which is a recurrence relation for the Fibonacci sequence and T^2-T-1=0 is the characteristic equation.
The real-valued solution is approx 2.15651 The other solution, which is complex-valued, is also valid -2.15651 + 14.0788i (where you take the logarithm of (1 - sqrt(5))/2)
I guessed that the number would be around that. If X is 1, then the numbers on the left are too large. If X = 2, the same thing, but if X = 3, the right side is larger. That would imply a number between 2 and 3, but closer to 2. I was adding numbers in my head, so there was no way to come up with an exact number.
Oops, I mis-spoke…the 25^x term increases faster with increasing x, so the answer is going to be a bit bigger than 2 by first inspection…ok, second inspection.
Alternatively, you could divide through by 25^x, giving (16/25)^x + (20/25)^x = 1. Then let r = (4/5)^x. That gives r^2 + r = 1 or r^2 + r - 1 = 0. That has solutions r = (√5 - 1)/2 and (-√5 - 1)/2. Since taking logs of negative numbers leads to complex solutions, we can take r = (√5 - 1)/2 to give a real solution when we take logs of (4/5)^x = (√5 - 1)/2. This solution is x = ln((√5 - 1)/2) / ln(4/5) ≈ 2.1565, which corresponds to the solution found in the video. That is the case because the numerators for the two paths to the solution are 1/ɸ and ɸ, respectively, and the denominators are also reciprocals of each other. Note that log(1/x) / log(1/y) = log(x) / log(y).
I saw the video preview on the RUclips front page and made a point to solve the question before coming to watch this video. I got x = log(ɸ) / log(5/4). I had to try a few things before I discovered a working set of algebraic manipulations. What I came up with was not exactly short. I'm curious to see whether there's a more direct path to the answer than the one I found. Well, I'm back, and the solution shown was almost exactly the one I found, no shorter. The explanation as to why we should exclude the negative solution to the quadratic part of the problem could have been clearer. We know that we are raising the positive value, 5/4, to some power. We cannot get a negative answer when we exponentiate a positive base, so we need only to consider the positive answer.
a little short cut here is let p=5^x and q=4^x . you can re-write it to 0 = p^2-pq-q^2 . Then just plug in to quadratic formula . a=1, b= -q, c=-q^2. the answer is p=q . golden ratio.
I saw this question in the thumbnail and I loved it. The fact that just two lines of working in it suddenly transforms into an elementary quadratic is delightful.
Quick note: It's not necessary to use natural logarithms (ln's) here, as the explanation seems to imply. Logs to any base at all, including base 10, work just as well, since you're simply dividing one logarithm by another in the same base.
That is true, but it's very common in formal mathematics to always use the natural logarithm (base e). That's because it really is the most "natural" base, and it leads to often reducing equations to natural functions such as e^x, which is equal to both its derivative and integral . If you study enough mathematics and physics, you'll see that ln (natural logarithm) is always used by convention.
@@egreenbery Indeed, and since log_b(x) = (ln x)/ln(b), the ratio on the right hand side of the solution would cancel to a ratio involving natural logs. And who wants a solution involving log_10 or log_2 or log_426265357?
@@rickdesper agreed. In fact, you can also say that log_b(x) = log x / log b , where the logs on the right side use any (but equivalent) positive number base. I just noticed the log emoji is this: 🪵. It looks like the joke "Log" toy on Ren & Stimpy.
Maybe I was luckly and found it easy, but here was my thought process. Figuring things out: The first immediate thought I had was to divide it into groups partly because we have two pretty square numbers and because maybe it would show me some nice connections: 16^x + 20^x = 25^x (4^x)(4^x) + (4^x)(5^x) = (5^x)(5^x) Dividing by 5^x because I notice I can factorize the expression in regards to 4^x on the leftside afterwards, (4^x)((4/5)^x) + (4^x) = (5^x) (4^x)( ((4/5)^x) + 1) = (5^x) Wow okay, now I notice that the equation can be easily substituted if I divide by 5^x again (4/5)^x ( ((4/5)^x) + 1) = 1 Substituting: Let v=(4/5)^x. The equation then becomes, v ( v + 1) = 1 Solving the substituted equation, v: Expanding and rearranging, v^2 + v - 1 =0 Using the quadratic formula to solve for possible values of v: v_- = 1/2 (-1 - \sqrt(5)); v_+ = 1/2 (-1 + \sqrt(5)) I remember when I was in highschool I forgot to check if the solution satisfied the original substitution, so I better check now. Looking at it, I see that v cannot be negative since v=(4/5)^x always greater than 0, and v_- is a negative number. Thus I have seen that v_- is not a legitimate solution to the original problem. v = 1/2 (-1 + \sqrt(5)) = (4/5)^x Solving for the original problem, x: v = 1/2 (-1 + \sqrt(5)) = (4/5)^x Taking the logarithm: -ln(2) + ln(-1+\sqrt(5)) = x (ln(4) - ln(5)) Solving the equation above for x we get, x = ( -ln(2) + ln(-1+\sqrt(5)) )/( ln(4) - ln(5) )
2 года назад+1
Great! Only there is one extra backslash in the last formula (before sqrt). Probably a typo. Or does it have some meaning?
@ yeah sorry that's how I write/initiate functions for the equation-field of Microsoft Word. 😅 Just means that sqrt isn't a bunch of variables multiplied together but is a function (the square root). I have a habit of writing it even when I don't need to. Sorry for the confusion 😃
Darn, I was working it correctly, but I missed one step. Back in 1991 I could have easily done it (that's when I earned my B.S. degree in mathematics) but 30 years later, I forgot so much.
I'm not quite that old, but I have easily forgotten how to use a generating question to derive the closed form expression of the Fibonacci sequence (only remember that it exists), but not stuff like this video's question
Nicely done, but I'd better do this : dividing by 20^x, I obtain (5/4)^x - (4/5)^x - 1 = 0... then replacing (5/4)^x by a, it gives a - 1/a - 1 = 0, or a^2 - a - 1 = 0. This Fibonaci equation gives the positive anxwer ɸ = (1+√5)/2, leading to the final answer x = ln (ɸ) / ln (5/4). Thanks to the authors !
@EE PI Of course, we can substitute the way you suggest, and reach the same solution. It's the other way to solve it fast and easy... Thanks for your remark
@EE PI I did the same, and it's worth noting that the solution is the same as the one in the video because one solution is log(1/ɸ)/log(1/1.25) which is the same as log(ɸ)/log(1.25) since log(1/a)/log(1/b) = log(a)/log(b).
It's not a distraction. It's a fundamental property of solutions of these sort of equations where you have (a.a)^x + (a.b)^x = (b.b)^x. Since you could choose to divide through by either a^2x or b^2x, giving solutions in terms of either t = (a/b)^x or t = (b/a)^x, it is necessary that the roots of the two resulting alternative polynomials (such as t^2 - t = 1 and t^2 + t = 1) are reciprocals of each other (for example the roots ɸ and ɸ-1, which is equal to 1/ɸ). This then works because log(1/ɸ)/log(a/b) = log(ɸ)/log(b/a).
The key to solvability of this, turns out to be that 16, 20, 25 are in geometric sequence. Any three (different!) positive numbers that are in geometric sequence, would, in place of 16, 20, 25, lead to a solution by this method. Slightly simpler, after dividing both sides by 16ˣ, group the fractions like this: 1 + (20/16)ˣ = (25/16)ˣ then do the reduction on the LHS and re-express the RHS: 1 + (5/4)ˣ = (5²/4²)ˣ = (5/4)²ˣ = [(5/4)ˣ]² and it's on to the quadratic in (5/4)ˣ, leading to the solution. Fred
No, that's not exactly the key. The method generally depends on the fact that 16, 20 and 25 are multiples of only two factors, in this case 4 and 5. I could ask 9^x + 33^x = 121^x and solve it the same way because the two factors are 3 and 11. Let u = (11/3)^x and then it's the same quadratic 1 + u = u^2. The only difference is that it's log(11/3) in the denominator of the answer.
@@davidquinn9676 Gottenhimfella is right. Look at what you're saying - the way you use those two factors (call them a and b) to produce the three numbers (that get raised to the "x" power), necessarily make them a geometric sequence of positive numbers: a², ab, b². And in fact, every geometric sequence of three positive numbers can be formed this way. The only further thing to add is that a and b, while necessarily different, can be any two unequal non-zero numbers of the same sign; they needn't be integers, or even rational. So I maintain that what I said is exactly the key.
1 + (5/4)^x = (5/4)^(2x) Writing y for (5/4)^x one gets y^2 = y + 1 or y = (1+√5)/2 , (1-√5)/2 The number (1+√5)/2 being the golden ratio one gets Hereby (5/4)^x = phi, - 1/phi or x = log (phi) to the base (5/4) Is the only fessible solution in real domain
Much simpler, from 16^x + 20^x + 25^x => 1 + (20/16)^x = (25/16)^x = (5/4)^2x =: phi², so: 1 + phi = phi² which defines the Golden Ratio phi(+-) = (sqrt(5)+-1)/2, thus (5/4)^x = phi, or x = log phi / log(5/4) -- two solutions which differ just in sign because phi(-)= 1/phi(+). Takes less than 30 seconds....
Pretty interesting case and straightforward solution, I like it! But I don't think you can pronounce "x - y" as "x negative y". It needs to be "x minus y". "Negative" is an adjective, it only describes the number. "Minus" is kind of like a verb, it means the action of taking away. :)
Just my 50 cents: You can find similar equations here on youtube. All are solved the same way: Divide by the highest power, to get rid of one x, substitude the power in such a way you get a polynomial. Solve the polynomial, pick the positive solution and re-substitude by taking the logs accordingly... Voila! Anyway... once you got rid of the highest power by dividing with it, the rest is quite straight forward. EDIT: you can find 3^x+9^x=27^x or 4^x+6^x=9^x here on YT. All solved the same way.
The equation t^2 - t - 1 = 0 is the most ubiquitous quadratic we meet in puzzle solving, and the root with the positive value - phi = the golden ratio.
My dear it is much easier; 1- reduce 25 to the power of x from both sides. 2-now log of the equation is; xlog16+xlog20-xlog25=0 Or; x[log16+log20-log25]=0 so as [log16+log20-log25] is not equal to zero. Then x must be zero X=0
x log 16 + x log 20 = x(log 16 + log 25) = x log(16x25) = log ((16x25)^x) = log((16^x)( 20^x)) Taking anti-logs gives (16^x)(20^x) which is not the same as the original (16^×) + (20^x)
I expressed the relation as y^2 + yz - z^2 = 0, where y = 4^x and z = 5^x. Solving this as a polynomial equation of degree 2 in the y variable gives the solution y/z = (1 + sqrt(5))/2 and by back-substitution and using the ln function on both sides we get x(ln(4/5)) = ln((1+sqrt(5))/2) which gives easily the answer after a division with ln(4/5).
This can be made shorter and less painful. Write: 4^(2x)+4^x*5^x=5^(2x). Divide by 4^x*5^x to get (4/5)^x+1=(5/4)^x. Thus (5/4)^x is the golden mean. Take logs and you are done.
"39 easy steps to get an answer that doesn't make sense". And this is why I never made it past Algebra. I'm glad that there are people in the world who have an affinity for mathematics.
@@cigmorfil4101 I used a movie reference as a bit of a non sequitur. "The 39 Steps" is a famous Alfred Hitchcock movie from 1935. A good mystery thriller and a must for any Hitchcock fan.
7:40 "However you can use your calculators and find out what exactly is this value." No you can't, because it's irrational. Some calculators maight give it in an irrational form, and I would write it as: log(base1.25)[(1+sqrt5)/2] from 6:42
He just shoud say "find out a decimal approximation" instead of "what exactly is this value", because the exact value is precisely the expression given with radicals...
An easy straightforward problem, solved in a minute or so. Tried to watch the explanation later to see if he had a different solution. Turns out he didn't, but I had to fast forward as the explanation was excruciatingly slow and he was trying to explain every dot produced on the piece of paper. Good he was not trying to explain what purpose the notepad or the marker were serving. Also, there is no arithmetic function called "NEGATIVE". The four basic functions are called ADDITION, SUBTRACTION, MULTIPLICATION and DIVISION. That is it. Sometimes people use shortcuts and say 'a PLUS b' or 'a MINUS b' but NO ONE says 'a NEGATIVE b'. One can say 'a PLUS NEGATIVE b', i.e., a+(-b)=a-b, but - again - NO ONE says "a NEGATIVE b".
can you please provide reference to back you claim instead of "NO ONE says" rational. Along the lines, you mentioned about- "the four basic functions", are you sure functions can also perform Operations?
@@mathed Yes, they are operations and I should have been more precise and said so. As for "no one says" - I still stand by it (maybe not in upper-case letters). Since there is no universal way to prove that a general statement is correct, one can challenge and disprove it by providing only one contrary example. Hence, to disprove my general (negative) statement "no one says that", all you have to do is to point me to one single positive example, e.g., math book that uses your language and says "a negative b". I'll be happy to correct myself if you do that - but, as that is a very long shot, I wouldn't hold my breath. Cheers.
@@mathed By the way, here is a gift for you. The answer could be simplified even more if you notice that you have multiples of 2 both in the numerator and denominator and choose log base 2 (since the base is irrelevant), not 10 or e or anything else. (Let's use notation log2 for it.) That will eliminate all multiples of 2 in the solution, as log2(2)=1 and log2(4)=2. In that case the solution will simply become (log2(1+sqrt(5))-1)/(log2(5)-2). Mathematically, the expression yields the same, but it doesn't have any fractions in the numerator and denominator - and in the place I come from that was considered bad math manners. In case anyone is interested in the numerical value of the expression, here is it to the 10th digit after decimal x = 2.1565123537.
@@am-gobears5191 Would you consider log(ɸ)/log(1.25) to be even better math manners? I doubt you can write the solution in simpler form (and the base of the log can be anything)
@@RexxSchneider Thank you very much, Rex - most certainly, the way you wrote it is more elegant. The reason I didn't go there was that I wasn't quite confident most folks would know what ɸ = 1/2(1+sqrt(5)) was - but, of course, I could have noted the definition. Having said that, it could actually be simplified even further. Since the base of the logarithm is irrelevant, nothing prevents us from choosing it to be 1.25, hence the answer would simply become log(ɸ). 🙂
You can get some pretty funny alternate forms by rationalizing the numerator or denominator of a fraction. Equivalently, treating either 4^x or 5^x as the "variable" in the quadratic substitution. Log identities also interact with these in various ways. I was concerned that maybe I made a mistake at first.
Yes, my solution was ln ((sqrt(5)-1)/2) / ln (4/5), which does work. Because I went for y=4^x/5^x for my substitution, so I have inverse values inside my logarithms.
If you don't limit yourself to real numbers, then you may take the second solution of the quadriatic equation as well: t = - 1/φ , which gives you the complex solution: x = - 2.1565 + 14.0788 i , which also satisfies the given equation.
...and if you limit yourself to natural integers, then you'll be sure at first view that no solution is possible (there is a truly marvelous proof of this, which this margin is too narrow to contain...)
@@Faxbable Yes I have because it was incomplete. You may see the solution to this problem (given by the clip) as an inherent proof to your claim. You just have to show that x given by: x = lnφ/(ln5-ln2) is not a natural integer.
@@shmuelzehavi4940 i wasn't serious... Joke is historical anecdote that refers to Fermat and the Fermat last theorem. Nevertheless, no need to solve this given equation, if you just want to point out there is no integer solution 🤓: apply the theorem (and check x = 1 & 2 don't work).
@@Faxbable I remember learning about it in 11th grade. Then, how shocked I was when I heard, less than a decade later than an American finally solved it (via an extremely complex proof).
I remember an iterative method called Runge-Kutha that I learned at my university, algebraic methods are not always easy for exponential equations, this is a very special case. Another possibility is a graphic method to check for multiple answers by the way I am not clear why one of the answers of the quadratic equation was discarded it would be conducent to a complex answer which is a valid answer anyway.
Yes you are right, after more than 45 years of me having the Numerical Methods course I made the mistake using the wrong iterative method name. Again you are right, Thank you.
What is called 'algebraic laws' are not laws at all but agreements on how to write things down. Algebra is nothing but different kinds of writing something.
A small point. The correct terminology is that the unary - operator is called negative and the binary - operator is called minus, so 5 - 3 is five minus three and -3 is negative three.
My approach: let a=4^x, b=5^x substitution: a^2+ab=b^2 quadratic formula: a= (-b +- sqrt(5b^2))/2 = b*(1+-sqrt(5))/2 substitution: 4^x=5^x * (1+-sqrt(5))/2 Division: .8^x=1+sqrt(5)/2 And then I knew the rest was trivial with logs and stopped.
Another way would be to divide both sides by 25expx. then let u = (4/5)expx--------------> quadratic equation u squared + u = 1 then u squared + u - 1 = 0. Solve by quadratic formula and then the rest is too lengthy, I don't want to talk anymore.
This is brilliant! I was working on a similar problem earlier, it’s pretty clear that any equation of the form a^x + b^x = c^x can be reduced down to an equation of the form a^x + b^X = 1. Is there a more general morbid of solving this when a^x isn’t a square of b^x? Specifically, I was looking at 4^x - 3^x >= 1, it’s pretty obvious that from this x>= 1 for any integer solutions of x, but it made me question of there is a verbal way of showing this algebraically.
A specific case of Fermat's Last Theorem, solvable if one base is product of square roots of other two. Negative infinity is also a "solution" of sorts,
Nothing to do with FLT. Here, x may be any real, so it is not necessary for one base to be the product of square roots of other two. We could for example have 3^x + 5^x = 7^x.
🤔 the spot I couldn't get passed was around 2:04 ... you wrote 16^x=(4^2)^x=4^2x. But then wrote that as 4^x×4^x, because of the formula you wrote just below, but the formula you used doesn't translate from the example given? 4^2x isn't 4^2+x. I'm not sure if that was an error or if the formula itself was written differently than it was meant to be written, but it seems like that's the basis for the argument made in finding the solution and I can't look past this point because I feel like the method used might be an incorrect example? One that just so happened to give the correct answer, despite the method not being a correct method? I'll admit it's been some time since I've studied algebra, so I'm very rusty on specific rules, but I was hoping for some clarification if possible?
Wow, I must be among the 3 percent of people who can solve this equation! Real solution: x = ( log(sqrt(5) - 1) - log(2) ) / ( log(4) - log(5) ) Complex solutions: x = ( log(sqrt(5) + 1) - log(2) + iπ(2n + 1) ) / ( log(4) - log(5) ) where n is an integer
Here's a similar problem that I created myself a while back as an exercise for whoever is reading my comment: 9^x + 27^x + 81^x = ( sin²(x)(1 + cot²(x)) + 2 )^x
reminds me of my math professor cranking the transparent film on the overhead projector forward line after line :-) but yes, well explained even if it exceeds my capabilities for a moment :-)
x = ln(golden ratio)/ln(5/4). Once you've done one, they can be solved mechanically. ln(golden ratio) upstairs and ln of fraction (second term/first term) in the basement.
Having in mind that 99% of students are not working towards a BSc in Maths, the usual Engineering or Technical students should be motivated to WORK OUT the final answer x= 2.156 to the very end, and actually try it replacing in the initial equation. One thing they would see is the precision obtained with 3 decimal places.
My problem there is that the ‘worked out form’ is by definition an approximation: the answer is irrational, so decimals are going to be ‘wrong’ no matter how precise you are. We do maths students a disservice by conflating approximate precision (a useful idea for engineers) and correctness (only possible by avoiding decimals altogether)
A difficult question? Maybe for someone in the elementary school. Honestly anyone with a bit of knowledge of algebra can solve the equation in 30sec., at least I did so---you only need to know that 16 and 25 are the squares of 4 and 5 and that 20=4X5.
I noticed you left out the +/- term. My solution kept it in, because it wasn't obvious that only real solution were wanted. If you are willing to go complex, logarithms and exponentials deliver all sorts of strange sinusoidal results.
I was born w/o a maths gene, so this squiggly stuff is sci-fi. As a matter of interest though, a couple of lines of Excel macro plus the Goal Seek function do the job. Just insert the numbers and click: x = 2.156513. :)
Easy.. Rewrite to 4^(2x)+4^x*5^x = 5^(2x) and then divide this equation by 4^x*5^x and use substitution y =(5/4)^x and solve quadratic equation 1/y+y=y^2. Solution is y=(1+5^0.5)/2 and then x = ln(y)/(ln5-ln4).
I've got a background in maths, but honestly, none of this stuff matters anymore. Get a grip, temperatures are soaring and species declining exponentially.
(a*b)^x = (a^x)*(b^x). its the same as if x is a number: (a*b)^2 = (a^2)*(b^2). He used (a^m)^n wich doesnt suit in this situation because we have 4 and 5 that corresponds to a and b. hope that helps
WOW..... FANTASTIC!. the result is x = 2.156512354 and if you substitute x with this value the result is ASTONISHING! not exactly but extremely close 16^(2.156..) + 20^(2.156...) = 25 ^ (2.156) 1,034.3675770052 = 1,034.367577098 and if you deduct both the difference will be only 0.0000000928 this is so precise though not perfect.
You did not really make clear why you threw out the negative answer for t. The reason is that t = (5/4)^x , and if x is a real number then that means that t MUST be positive, since any positive real number base raised to a real number power is positive. (Of course, if you are NOT assuming that x is a real number, then you WOULD need to also consider the negative value of t when solving for x.) It is worth noting also that this equation could be solved in a similar manner by first dividing through by either 20^x or 25^x, in lieu of 16^x.
Why divide by 16 power of x in the first place? How you think about going about solving a problem is as important as the mechanics of the laws of exponents and logs.
The -b+/-sqtr (b2 - 4ac)/2a as you can see at 4:58 in the video is known and used alot..... but what about the ruzzian method/algorithm/formula for the same purpose to find zero point?
I have not watched more than 30 seconds of this video. From a brief glance at the equation it is obvious that the value of x is very close to 2. A few seconds work on with a calculator shows that that the answer lies between 2.1 and 2.2
@@geoninja8971 Thank you! I have now seen it from beginning to end. What I saw was a laborious demonstration of how to solve one very specific problem. The numbers 16, 20 and 25 allow the equation to be re-written as a quadratic: other numbers would not.
I 'solved' it using excel. Put -25^x on the left side so it was a zero answer, then kept finding the zero crossover point and added another decimal to the series. Finally ended up with 2.15651235370905. Here are the final attempts around zero... 2.15651235370904=4.54747E-12 2.15651235370905=0 2.15651235370906=-2.50111E-12
@@dougawhite Using a numerical method (trial and error) to solve. Using the newton raphson method starting with x{0}=2 you get the approximate numerical answer in 7 iterations (using excel before it has reached the limit of its fp representation): 2 2.29887650904642 2.19959005179848 2.16150847084116 2.15658726823254 2.15651237079779 2.15651235370906 2.15651235379905 (In fact it bounces between ...05 and ...06 as the limitations of the rounded display vs the limitations of the internal (binary) representation is evidenced - the difference is in the 15th decimal place which is used to round the displayed digits) Let f(x) = 16^x + 20^x - 25^x Then f'(x) = (ln 16)16^x + (ln 20)20^x - (ln 25)25^x And x{r+1} = x{r} - f(x{r})/f'(x{r}) Select a reasonable guess for x0. (Though personally I wouldn't use excal for this kind of things as I've heard it uses it's own weird fp mumber representations and algorithms, not IEEE ones.)
Excellent explanation and well set out exact solution
5square _ 4 square = 20 to the poer x
5+4 × 5- 4 = 20 to the poer x from that x= 1
Divide and multiply, divide and multiply, SUBTRACT the length of your arm. They don't like it up 'em.
Really?
I recommend this maths problem . This guy is crazy applied a method that I have seen before in RUclips .
ruclips.net/video/z2OyVIJznHw/видео.html
anyone else noticed that T is the golden ratio because R(n+2)=R(n+1)+R(n) which is a recurrence relation for the Fibonacci sequence and T^2-T-1=0 is the characteristic equation.
@Aristotle was Not a fan of Plato mai
S
Yes 😀👍🏽
NICE!
What did/do you study?
Nice! thank you!
The real-valued solution is approx 2.15651
The other solution, which is complex-valued, is also valid -2.15651 + 14.0788i (where you take the logarithm of (1 - sqrt(5))/2)
Congrats for an excellent actual answer!
You mean the log[(√(5) -1)/2] / log(4/5) giving you real valued 2.15651..
which is simply the solution to (4/5)^x+1= 1/(4/5)^x
but well spotted.
I guessed that the number would be around that. If X is 1, then the numbers on the left are too large. If X = 2, the same thing, but if X = 3, the right side is larger. That would imply a number between 2 and 3, but closer to 2. I was adding numbers in my head, so there was no way to come up with an exact number.
Oops, I mis-spoke…the 25^x term increases faster with increasing x, so the answer is going to be a bit bigger than 2 by first inspection…ok, second inspection.
@EE PI Great, this is the answer i found, when i calculated it in my way. It is also the answer i can understand.
Thank you for this thought-provoking question. Once the division of 16^x was carried out, the solution was rather straightforward. Well done.
Alternatively, you could divide through by 25^x, giving (16/25)^x + (20/25)^x = 1. Then let r = (4/5)^x. That gives r^2 + r = 1 or r^2 + r - 1 = 0. That has solutions r = (√5 - 1)/2 and (-√5 - 1)/2. Since taking logs of negative numbers leads to complex solutions, we can take r = (√5 - 1)/2 to give a real solution when we take logs of (4/5)^x = (√5 - 1)/2.
This solution is x = ln((√5 - 1)/2) / ln(4/5) ≈ 2.1565, which corresponds to the solution found in the video. That is the case because the numerators for the two paths to the solution are 1/ɸ and ɸ, respectively, and the denominators are also reciprocals of each other. Note that log(1/x) / log(1/y) = log(x) / log(y).
I saw the video preview on the RUclips front page and made a point to solve the question before coming to watch this video. I got x = log(ɸ) / log(5/4). I had to try a few things before I discovered a working set of algebraic manipulations. What I came up with was not exactly short. I'm curious to see whether there's a more direct path to the answer than the one I found.
Well, I'm back, and the solution shown was almost exactly the one I found, no shorter.
The explanation as to why we should exclude the negative solution to the quadratic part of the problem could have been clearer. We know that we are raising the positive value, 5/4, to some power. We cannot get a negative answer when we exponentiate a positive base, so we need only to consider the positive answer.
a little short cut here is let p=5^x and q=4^x . you can re-write it to 0 = p^2-pq-q^2 . Then just plug in to quadratic formula . a=1, b= -q, c=-q^2. the answer is p=q . golden ratio.
Dude! Sweet move.
yeah, this solution is actually very elegant, and related to the golden ratio. a pity the video does not mention that
I saw this question in the thumbnail and I loved it. The fact that just two lines of working in it suddenly transforms into an elementary quadratic is delightful.
Quick note: It's not necessary to use natural logarithms (ln's) here, as the explanation seems to imply. Logs to any base at all, including base 10, work just as well, since you're simply dividing one logarithm by another in the same base.
That is true, but it's very common in formal mathematics to always use the natural logarithm (base e). That's because it really is the most "natural" base, and it leads to often reducing equations to natural functions such as e^x, which is equal to both its derivative and integral . If you study enough mathematics and physics, you'll see that ln (natural logarithm) is always used by convention.
Base e is, um, the natural choice for logarithm.
I'll show myself out.
@@egreenbery Indeed, and since log_b(x) = (ln x)/ln(b), the ratio on the right hand side of the solution would cancel to a ratio involving natural logs. And who wants a solution involving log_10 or log_2 or log_426265357?
@@rickdesper
Cute. Very cute.
The door is over there.
@@rickdesper agreed. In fact, you can also say that log_b(x) = log x / log b , where the logs on the right side use any (but equivalent) positive number base.
I just noticed the log emoji is this: 🪵. It looks like the joke "Log" toy on Ren & Stimpy.
Maybe I was luckly and found it easy, but here was my thought process.
Figuring things out:
The first immediate thought I had was to divide it into groups partly because we have two pretty square numbers and because maybe it would show me some nice connections:
16^x + 20^x = 25^x
(4^x)(4^x) + (4^x)(5^x) = (5^x)(5^x)
Dividing by 5^x because I notice I can factorize the expression in regards to 4^x on the leftside afterwards,
(4^x)((4/5)^x) + (4^x) = (5^x)
(4^x)( ((4/5)^x) + 1) = (5^x)
Wow okay, now I notice that the equation can be easily substituted if I divide by 5^x again
(4/5)^x ( ((4/5)^x) + 1) = 1
Substituting:
Let v=(4/5)^x. The equation then becomes,
v ( v + 1) = 1
Solving the substituted equation, v:
Expanding and rearranging,
v^2 + v - 1 =0
Using the quadratic formula to solve for possible values of v:
v_- = 1/2 (-1 - \sqrt(5));
v_+ = 1/2 (-1 + \sqrt(5))
I remember when I was in highschool I forgot to check if the solution satisfied the original substitution, so I better check now. Looking at it, I see that v cannot be negative since v=(4/5)^x always greater than 0, and v_- is a negative number. Thus I have seen that v_- is not a legitimate solution to the original problem.
v = 1/2 (-1 + \sqrt(5)) = (4/5)^x
Solving for the original problem, x:
v = 1/2 (-1 + \sqrt(5)) = (4/5)^x
Taking the logarithm:
-ln(2) + ln(-1+\sqrt(5)) = x (ln(4) - ln(5))
Solving the equation above for x we get,
x = ( -ln(2) + ln(-1+\sqrt(5)) )/( ln(4) - ln(5) )
Great! Only there is one extra backslash in the last formula (before sqrt). Probably a typo. Or does it have some meaning?
@ yeah sorry that's how I write/initiate functions for the equation-field of Microsoft Word. 😅
Just means that sqrt isn't a bunch of variables multiplied together but is a function (the square root). I have a habit of writing it even when I don't need to. Sorry for the confusion 😃
Amazing, just continue !👏🤩
Darn, I was working it correctly, but I missed one step. Back in 1991 I could have easily done it (that's when I earned my B.S. degree in mathematics) but 30 years later, I forgot so much.
You think you have trouble remembering? 🤣 I completed my BSc study in 1975.
@@brianvogt8125 @bjbell52 I can relate so much to both of you. 1983 for me. At least I have some great memories.
I'm not quite that old, but I have easily forgotten how to use a generating question to derive the closed form expression of the Fibonacci sequence (only remember that it exists), but not stuff like this video's question
Does B.S. mean Bachelor of Shit ? Does B.Sc. mean Bachelor of Scum? DIVIDE AND MULTIPLY! DIVIDE AND MULTIPLY! SUBTRACT THE LENGTH OF YOUR ARM!
@@brianvogt8125 81 lol. but some i still remember! EE here.
Nicely done, but I'd better do this : dividing by 20^x, I obtain (5/4)^x - (4/5)^x - 1 = 0... then replacing (5/4)^x by a, it gives a - 1/a - 1 = 0, or a^2 - a - 1 = 0.
This Fibonaci equation gives the positive anxwer ɸ = (1+√5)/2, leading to the final answer x = ln (ɸ) / ln (5/4). Thanks to the authors !
Thanks for your appreciation... please go on with these tricky exercises !
Did it exactly this way without looking at the video...well done!
@@leonardo.molinari69 Thanks, we understand each-other... Have a nice time !
@EE PI Of course, we can substitute the way you suggest, and reach the same solution. It's the other way to solve it fast and easy... Thanks for your remark
@EE PI I did the same, and it's worth noting that the solution is the same as the one in the video because one solution is log(1/ɸ)/log(1/1.25) which is the same as log(ɸ)/log(1.25) since log(1/a)/log(1/b) = log(a)/log(b).
Divide on both sides by 25^x, and set T = (4/5)^x. The new equation is T^2 + T = 1, or T^2 + T - 1 = 0. Now solve the quadratic, and then solve for x.
I did it that way :-)
@@GamleMich me too
Is the answer the same?
a=4^x, b=5^x. Solve equation b*b-ab-a*a=0 (second order). Result b=((1-sqrt(5))/2)*a, so you have b/a. Quite straightforward.
OR, you can *solve the given equation experimentally* a digit at a time, by numerical methods.
Impressed that he resisted the distraction of mentioning the Golden Ratio
It's not a distraction. It's a fundamental property of solutions of these sort of equations where you have (a.a)^x + (a.b)^x = (b.b)^x.
Since you could choose to divide through by either a^2x or b^2x, giving solutions in terms of either t = (a/b)^x or t = (b/a)^x, it is necessary that the roots of the two resulting alternative polynomials (such as t^2 - t = 1 and t^2 + t = 1) are reciprocals of each other (for example the roots ɸ and ɸ-1, which is equal to 1/ɸ). This then works because log(1/ɸ)/log(a/b) = log(ɸ)/log(b/a).
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The key to solvability of this, turns out to be that 16, 20, 25 are in geometric sequence.
Any three (different!) positive numbers that are in geometric sequence, would, in place of 16, 20, 25, lead to a solution by this method.
Slightly simpler, after dividing both sides by 16ˣ, group the fractions like this:
1 + (20/16)ˣ = (25/16)ˣ
then do the reduction on the LHS and re-express the RHS:
1 + (5/4)ˣ = (5²/4²)ˣ = (5/4)²ˣ = [(5/4)ˣ]²
and it's on to the quadratic in (5/4)ˣ, leading to the solution.
Fred
ooh. it is leading to 'new' quadratic equation based on (5/4)^x.
so, i think NOT all number can be put on power x there, right ?
Yes, I thought it was misleading of him not to point out that this was a prerequisite
No, that's not exactly the key. The method generally depends on the fact that 16, 20 and 25 are multiples of only two factors, in this case 4 and 5. I could ask 9^x + 33^x = 121^x and solve it the same way because the two factors are 3 and 11. Let u = (11/3)^x and then it's the same quadratic 1 + u = u^2. The only difference is that it's log(11/3) in the denominator of the answer.
@@davidquinn9676 but that amounts to the same thing. The example you gave is also a geometric progression, ratio 3.666...
@@davidquinn9676 Gottenhimfella is right. Look at what you're saying - the way you use those two factors (call them a and b) to produce the three numbers (that get raised to the "x" power), necessarily make them a geometric sequence of positive numbers: a², ab, b².
And in fact, every geometric sequence of three positive numbers can be formed this way.
The only further thing to add is that a and b, while necessarily different, can be any two unequal non-zero numbers of the same sign; they needn't be integers, or even rational.
So I maintain that what I said is exactly the key.
1 + (5/4)^x = (5/4)^(2x)
Writing y for (5/4)^x one gets
y^2 = y + 1
or y = (1+√5)/2 , (1-√5)/2
The number (1+√5)/2 being the golden ratio one gets
Hereby (5/4)^x = phi, - 1/phi
or x = log (phi) to the base (5/4)
Is the only fessible solution in real domain
Quite elegant. Thanks for sharing.
Much simpler, from 16^x + 20^x + 25^x => 1 + (20/16)^x = (25/16)^x = (5/4)^2x =: phi², so: 1 + phi = phi² which defines the Golden Ratio phi(+-) = (sqrt(5)+-1)/2, thus (5/4)^x = phi, or x = log phi / log(5/4) -- two solutions which differ just in sign because phi(-)= 1/phi(+). Takes less than 30 seconds....
Pretty interesting case and straightforward solution, I like it! But I don't think you can pronounce "x - y" as "x negative y". It needs to be "x minus y". "Negative" is an adjective, it only describes the number. "Minus" is kind of like a verb, it means the action of taking away. :)
Correct. "x minus y", "minus y", "minus one", "a negative number". But the verb is "subtract". "Subtract y from x".
Likewise "positive " and "plus "
[Klaxons sound. A voice crackles through overhead speakers: *Warning! Warning! English majors have breached the Maths lab...*]
it's called "text to speech"
@@richardgale4827 Or... people who studied maths in an English-speaking country.
Just my 50 cents: You can find similar equations here on youtube. All are solved the same way: Divide by the highest power, to get rid of one x, substitude the power in such a way you get a polynomial. Solve the polynomial, pick the positive solution and re-substitude by taking the logs accordingly... Voila!
Anyway... once you got rid of the highest power by dividing with it, the rest is quite straight forward.
EDIT: you can find 3^x+9^x=27^x or 4^x+6^x=9^x here on YT. All solved the same way.
The equation t^2 - t - 1 = 0 is the most ubiquitous quadratic we meet in puzzle solving, and the root with the positive value - phi = the golden ratio.
And the other solution is the golden ratio's little brother.
My dear it is much easier;
1- reduce 25 to the power of x from both sides.
2-now log of the equation is;
xlog16+xlog20-xlog25=0
Or;
x[log16+log20-log25]=0 so as [log16+log20-log25] is not equal to zero.
Then x must be zero X=0
x log 16 + x log 20
= x(log 16 + log 25)
= x log(16x25)
= log ((16x25)^x)
= log((16^x)( 20^x))
Taking anti-logs gives
(16^x)(20^x) which is not the same as the original (16^×) + (20^x)
I expressed the relation as y^2 + yz - z^2 = 0, where y = 4^x and z = 5^x. Solving this as a polynomial equation of degree 2 in the y variable gives the solution y/z = (1 + sqrt(5))/2 and by back-substitution and using the ln function on both sides we get x(ln(4/5)) = ln((1+sqrt(5))/2) which gives easily the answer after a division with ln(4/5).
Same here!
@@thomassimpson9886 Ditto! V. pleased with myself, as it avoided complications and could all be done in my head.
This can be made shorter and less painful. Write: 4^(2x)+4^x*5^x=5^(2x). Divide by 4^x*5^x to get (4/5)^x+1=(5/4)^x. Thus (5/4)^x is the golden mean. Take logs and you are done.
"39 easy steps to get an answer that doesn't make sense". And this is why I never made it past Algebra. I'm glad that there are people in the world who have an affinity for mathematics.
No, sorry, The 39 Steps is something entirely different. 😅
Why does the answer not make sense?
@@LesCish
Even then the answer varies depending upon whether you use the original or not.
@@violjohn
Because Hollywood likes to mess things up and make them it's own.
@@cigmorfil4101 I used a movie reference as a bit of a non sequitur. "The 39 Steps" is a famous Alfred Hitchcock movie from 1935. A good mystery thriller and a must for any Hitchcock fan.
7:40 "However you can use your calculators and find out what exactly is this value." No you can't, because it's irrational. Some calculators maight give it in an irrational form, and I would write it as:
log(base1.25)[(1+sqrt5)/2] from 6:42
He just shoud say "find out a decimal approximation" instead of "what exactly is this value", because the exact value is precisely the expression given with radicals...
WOW Amazing !!!
Just came for math videos and I came across your video and I said I've to watch it (It's worth it) thanks
An easy straightforward problem, solved in a minute or so.
Tried to watch the explanation later to see if he had a different solution. Turns out he didn't, but I had to fast forward as the explanation was excruciatingly slow and he was trying to explain every dot produced on the piece of paper. Good he was not trying to explain what purpose the notepad or the marker were serving.
Also, there is no arithmetic function called "NEGATIVE". The four basic functions are called ADDITION, SUBTRACTION, MULTIPLICATION and DIVISION. That is it. Sometimes people use shortcuts and say 'a PLUS b' or 'a MINUS b' but NO ONE says 'a NEGATIVE b'. One can say 'a PLUS NEGATIVE b', i.e., a+(-b)=a-b, but - again - NO ONE says "a NEGATIVE b".
can you please provide reference to back you claim instead of "NO ONE says" rational. Along the lines, you mentioned about- "the four basic functions", are you sure functions can also perform Operations?
@@mathed Yes, they are operations and I should have been more precise and said so. As for "no one says" - I still stand by it (maybe not in upper-case letters). Since there is no universal way to prove that a general statement is correct, one can challenge and disprove it by providing only one contrary example. Hence, to disprove my general (negative) statement "no one says that", all you have to do is to point me to one single positive example, e.g., math book that uses your language and says "a negative b". I'll be happy to correct myself if you do that - but, as that is a very long shot, I wouldn't hold my breath.
Cheers.
@@mathed By the way, here is a gift for you. The answer could be simplified even more if you notice that you have multiples of 2 both in the numerator and denominator and choose log base 2 (since the base is irrelevant), not 10 or e or anything else. (Let's use notation log2 for it.) That will eliminate all multiples of 2 in the solution, as log2(2)=1 and log2(4)=2. In that case the solution will simply become (log2(1+sqrt(5))-1)/(log2(5)-2). Mathematically, the expression yields the same, but it doesn't have any fractions in the numerator and denominator - and in the place I come from that was considered bad math manners.
In case anyone is interested in the numerical value of the expression, here is it to the 10th digit after decimal x = 2.1565123537.
@@am-gobears5191 Would you consider log(ɸ)/log(1.25) to be even better math manners? I doubt you can write the solution in simpler form (and the base of the log can be anything)
@@RexxSchneider Thank you very much, Rex - most certainly, the way you wrote it is more elegant. The reason I didn't go there was that I wasn't quite confident most folks would know what ɸ = 1/2(1+sqrt(5)) was - but, of course, I could have noted the definition. Having said that, it could actually be simplified even further. Since the base of the logarithm is irrelevant, nothing prevents us from choosing it to be 1.25, hence the answer would simply become log(ɸ). 🙂
You can get some pretty funny alternate forms by rationalizing the numerator or denominator of a fraction. Equivalently, treating either 4^x or 5^x as the "variable" in the quadratic substitution. Log identities also interact with these in various ways. I was concerned that maybe I made a mistake at first.
x=0 😶
Yes, my solution was ln ((sqrt(5)-1)/2) / ln (4/5), which does work. Because I went for y=4^x/5^x for my substitution, so I have inverse values inside my logarithms.
If you don't limit yourself to real numbers, then you may take the second solution of the quadriatic equation as well: t = - 1/φ , which gives you the complex solution: x = - 2.1565 + 14.0788 i , which also satisfies the given equation.
...and if you limit yourself to natural integers, then you'll be sure at first view that no solution is possible (there is a truly marvelous proof of this, which this margin is too narrow to contain...)
i cannot read your reply, maybe you have deleted it?
@@Faxbable Yes I have because it was incomplete.
You may see the solution to this problem (given by the clip) as an inherent proof to your claim. You just have to show that x given by: x = lnφ/(ln5-ln2) is not a natural integer.
@@shmuelzehavi4940 i wasn't serious... Joke is historical anecdote that refers to Fermat and the Fermat last theorem.
Nevertheless, no need to solve this given equation, if you just want to point out there is no integer solution 🤓: apply the theorem (and check x = 1 & 2 don't work).
@@Faxbable I remember learning about it in 11th grade. Then, how shocked I was when I heard, less than a decade later than an American finally solved it (via an extremely complex proof).
I remember an iterative method called Runge-Kutha that I learned at my university, algebraic methods are not always easy for exponential equations, this is a very special case. Another possibility is a graphic method to check for multiple answers by the way I am not clear why one of the answers of the quadratic equation was discarded it would be conducent to a complex answer which is a valid answer anyway.
isn't runge-kutta for differential equations?
Yes you are right, after more than 45 years of me having the Numerical Methods course I made the mistake using the wrong iterative method name. Again you are right, Thank you.
Additionally, (1 + sqrt 5)/2 is simply 1/2 + (5/4)^(1/2) power, which is then equal to (5/4)^x. Using this we could use log base 5/4 if we wanted.
I wonder if a^x + b^x + c^x always has a solution for x for any choice of a, b, c? Or only for certain ones? (x may have to be complex?)
why do I feel triggered that he seems to be pushing down awfully hard with a felt-tip marker?
other than that, excellent, thorough explanation.
This is because of the index finger, which is placed at such a position, that it tends to force the pen hard on the paper.
What is called 'algebraic laws' are not laws at all but agreements on how to write things down. Algebra is nothing but different kinds of writing something.
Well done 👍🏻
A small point. The correct terminology is that the unary - operator is called negative and the binary - operator is called minus, so 5 - 3 is five minus three and -3 is negative three.
The unary - operator is called minus, too. "Minus y", "minus three". "Negative" is an adjective, e.g. "x is negative" or "a negative number".
(1+sqr5)/2 is actually equal to fi (golden ratio)
what the L9 gangmember doin here
@@kojixd uninstalled league now gigagrind calculus
Phi
Φ or φ (phi actually)
@@createyourownfuture3840 or the other way to write phi
My approach:
let a=4^x, b=5^x
substitution: a^2+ab=b^2
quadratic formula: a= (-b +- sqrt(5b^2))/2 = b*(1+-sqrt(5))/2
substitution: 4^x=5^x * (1+-sqrt(5))/2
Division: .8^x=1+sqrt(5)/2
And then I knew the rest was trivial with logs and stopped.
Another way would be to divide both sides by 25expx. then let u = (4/5)expx--------------> quadratic equation u squared + u = 1 then u squared + u - 1 = 0. Solve by quadratic formula and then the rest is too lengthy, I don't want to talk anymore.
When he said "you all know" I just thought "not me Master" just to confirm that a second later...
Sir why not log is used in this question. If log can't use, then please tell me that why. If we use log then question will be so easy
a problem can be solved in many ways, and you have the liberty to make the choice.
@@mathed sir but when we use log then answer is not coming that u are solved
Dear Mr. Purushottam, can you share your working here, I can check for you. Best
Divide by 20^x both sides and get a quadratic equation where variable is a=(4/5)^×
Obtain the result of a and then x.
It's simple and fast
This is brilliant! I was working on a similar problem earlier, it’s pretty clear that any equation of the form a^x + b^x = c^x can be reduced down to an equation of the form a^x + b^X = 1. Is there a more general morbid of solving this when a^x isn’t a square of b^x?
Specifically, I was looking at 4^x - 3^x >= 1, it’s pretty obvious that from this x>= 1 for any integer solutions of x, but it made me question of there is a verbal way of showing this algebraically.
👌 awesome!
This is nice problem to challenge good students.
the answer is so simple it jumped right out at me just from the thumbnail!!
X = X!!!!!! naturally
Nice to see the Golden Ratio appear in the answer. :)
I thought I recognized that, wasn't positive.
You've employed the Golden Ratio, which so intrigued the ancient Greeks!
x can further be defined as {ln[1 + (5)^2] - ln(2)/[ln(5) - ln(4)]
If you're an ethereal being, Mind Over Body is undefined.
Watch it there Elmer! Them th-airz fightin' words! lol
Do people even realize who I am? I'm the DOC HOLIDAY of math. I'm your huckleberry. That's JUST my game! :)
For a second I thought we was going to disprove Fermat's last theorem.
Brilliant explanation👌👌
A specific case of Fermat's Last Theorem, solvable if one base is product of square roots of other two. Negative infinity is also a "solution" of sorts,
Nothing to do with FLT. Here, x may be any real, so it is not necessary for one base to be the product of square roots of other two. We could for example have 3^x + 5^x = 7^x.
🤔 the spot I couldn't get passed was around 2:04 ... you wrote 16^x=(4^2)^x=4^2x. But then wrote that as 4^x×4^x, because of the formula you wrote just below, but the formula you used doesn't translate from the example given? 4^2x isn't 4^2+x. I'm not sure if that was an error or if the formula itself was written differently than it was meant to be written, but it seems like that's the basis for the argument made in finding the solution and I can't look past this point because I feel like the method used might be an incorrect example? One that just so happened to give the correct answer, despite the method not being a correct method? I'll admit it's been some time since I've studied algebra, so I'm very rusty on specific rules, but I was hoping for some clarification if possible?
I now have a migraine, thank you.
A lot of mysterious things can be reduced down to the quadratic formula.
As a teacher, as I presume you are, you should know that from graphical point of view the equal sign should be alined with fraction line.
I subscribe your point of view; I hate very much this misaligned way of writing!
So easy question. I solve this in my mind easily.
This also sounds so nice to American ears when you hear the lecturer's Indian/Pakistani accent (I can't tell the difference).
If you use the negative value of t, then ln ( negative) forbidden? or do you get imaginary exponentials?
Wow! the last maths problem I solved the answer was 'there was five apples left in the basket'.
Problems like this made me change my major to biology.
Wow, I must be among the 3 percent of people who can solve this equation!
Real solution: x = ( log(sqrt(5) - 1) - log(2) ) / ( log(4) - log(5) )
Complex solutions: x = ( log(sqrt(5) + 1) - log(2) + iπ(2n + 1) ) / ( log(4) - log(5) ) where n is an integer
Here's a similar problem that I created myself a while back as an exercise for whoever is reading my comment:
9^x + 27^x + 81^x = ( sin²(x)(1 + cot²(x)) + 2 )^x
OK, know it's a really old chestnut in math RUclips. I saw equation like this so many times :>
I did it !😀
Thanks for these challenges
I never knew how to do this calculation. Now I still don’t but now at least I know who to ask.
M
reminds me of my math professor cranking the transparent film on the overhead projector forward line after line :-) but yes, well explained even if it exceeds my capabilities for a moment :-)
x = ln(golden ratio)/ln(5/4). Once you've done one, they can be solved mechanically. ln(golden ratio) upstairs and ln of fraction (second term/first term) in the basement.
This can also be solved by dividing by 25 to the x power or 20 to the x power.
Divide by (5^x)^2 and set u = (4/5)^x, then u^2+u-1 =0 and the quadratic formula can be used. So x=ln((sqrt5-1)/2)/ln(4/5)
Best Wishes
Having in mind that 99% of students are not working towards a BSc in Maths, the usual Engineering or Technical students should be motivated to WORK OUT the final answer x= 2.156 to the very end, and actually try it replacing in the initial equation. One thing they would see is the precision obtained with 3 decimal places.
My problem there is that the ‘worked out form’ is by definition an approximation: the answer is irrational, so decimals are going to be ‘wrong’ no matter how precise you are.
We do maths students a disservice by conflating approximate precision (a useful idea for engineers) and correctness (only possible by avoiding decimals altogether)
A difficult question? Maybe for someone in the elementary school. Honestly anyone with a bit of knowledge of algebra can solve the equation in 30sec., at least I did so---you only need to know that 16 and 25 are the squares of 4 and 5 and that 20=4X5.
Really interesting. No one seems to have noticed tho that the value of t is the golden ratio phi?
I had the gist of the solution,but there were
many parts in arriving at the solution.Could
it not been done easy.
I noticed you left out the +/- term. My solution kept it in, because it wasn't obvious that only real solution were wanted. If you are willing to go complex, logarithms and exponentials deliver all sorts of strange sinusoidal results.
This was a great problem, and you presented a great solution.
Thank you.
I was born w/o a maths gene, so this squiggly stuff is sci-fi. As a matter of interest though, a couple of lines of Excel macro plus the Goal Seek function do the job. Just insert the numbers and click: x = 2.156513. :)
Wrong. 20 to the x is not equal to 4 to the x times 5 to the x. That is already equal to 20 to the 2x.
Neatly done.
Easy.. Rewrite to 4^(2x)+4^x*5^x = 5^(2x) and then divide this equation by 4^x*5^x and use substitution y =(5/4)^x and solve quadratic equation 1/y+y=y^2. Solution is y=(1+5^0.5)/2
and then x = ln(y)/(ln5-ln4).
I suspect that there is a second solution lurking in there somewhere.
I've got a background in maths, but honestly, none of this stuff matters anymore. Get a grip, temperatures are soaring and species declining exponentially.
I don't understand how 20^x = (5^x)(4^x). Which did you use please.?
a^n×b^n = (a*b)^n
(a*b)^x = (a^x)*(b^x). its the same as if x is a number: (a*b)^2 = (a^2)*(b^2).
He used (a^m)^n wich doesnt suit in this situation because we have 4 and 5 that corresponds to a and b. hope that helps
same as..... square root 100 (ie 10) = square root of 50 x square root of 2
knowing this rule will help simplifying equations with irrational numbers
Because 5 * 4 = 4 + 4 + 4 + 4 + 4 = successor(3) + successor(3) + successor(3) + successor(3) + successor(3) = successor(successor(2)) + successor(successor(2)) + successor(successor(2)) + successor(successor(2)) + successor(successor(2)) = successor(successor( successor(1)) + successor(successor( successor(1)) + successor(successor( successor(1)) + successor(successor( successor(1)) + successor(successor( successor(1)) = successor(successor(successor(successor(successor(successor(successor(successor(successor(successor(successor(successor(successor(successor(successor(successor(successor(successor(successor(1)))))))))))))))))) = 20, and thus 5 * 4 = 20 (that's how I learned it).
That was mind boggling. Turned out that X is a ridiculously complicated real number.
But it's elegant and fully solved by operation. If someone gave the answer by some certain number I'd be less impressed
WOW..... FANTASTIC!. the result is x = 2.156512354
and if you substitute x with this value
the result is ASTONISHING!
not exactly but extremely close
16^(2.156..) + 20^(2.156...) = 25 ^ (2.156)
1,034.3675770052 = 1,034.367577098
and if you deduct both the difference will be only 0.0000000928
this is so precise though not perfect.
You did not really make clear why you threw out the negative answer for t. The reason is that t = (5/4)^x , and if x is a real number then that means that t MUST be positive, since any positive real number base raised to a real number power is positive. (Of course, if you are NOT assuming that x is a real number, then you WOULD need to also consider the negative value of t when solving for x.)
It is worth noting also that this equation could be solved in a similar manner by first dividing through by either 20^x or 25^x, in lieu of 16^x.
Ooh a golden ratio solution!
Why divide by 16 power of x in the first place? How you think about going about solving a problem is as important as the mechanics of the laws of exponents and logs.
The -b+/-sqtr (b2 - 4ac)/2a as you can see at 4:58 in the video is known and used alot..... but what about the ruzzian method/algorithm/formula for the same purpose to find zero point?
The value, (1+sqrt(5))/2 is also known as the Golden Ratio. The Greeks knew about this.
Excellent
Well explained!
Don't be so proud, claiming 97% won't be able to... There is always someone better. One should then strive to be better.
the complex value of t is actually valid but if we only intrestet in reel solutions we not use it
I have not watched more than 30 seconds of this video. From a brief glance at the equation it is obvious that the value of x is very close to 2. A few seconds work on with a calculator shows that that the answer lies between 2.1 and 2.2
You should watch it, and learn something.... :)
@@geoninja8971 Thank you! I have now seen it from beginning to end. What I saw was a laborious demonstration of how to solve one very specific problem. The numbers 16, 20 and 25 allow the equation to be re-written as a quadratic: other numbers would not.
Finally one with common sense. They worked so hard and didn't even get an answer only a theory. I would have give just ~2 not even 2.1 thanks.
I 'solved' it using excel. Put -25^x on the left side so it was a zero answer, then kept finding the zero crossover point and added another decimal to the series. Finally ended up with 2.15651235370905. Here are the final attempts around zero... 2.15651235370904=4.54747E-12 2.15651235370905=0 2.15651235370906=-2.50111E-12
@@dougawhite
Using a numerical method (trial and error) to solve.
Using the newton raphson method starting with x{0}=2 you get the approximate numerical answer in 7 iterations (using excel before it has reached the limit of its fp representation):
2
2.29887650904642
2.19959005179848
2.16150847084116
2.15658726823254
2.15651237079779
2.15651235370906
2.15651235379905
(In fact it bounces between ...05 and ...06 as the limitations of the rounded display vs the limitations of the internal (binary) representation is evidenced - the difference is in the 15th decimal place which is used to round the displayed digits)
Let f(x) = 16^x + 20^x - 25^x
Then f'(x) = (ln 16)16^x + (ln 20)20^x - (ln 25)25^x
And x{r+1} = x{r} - f(x{r})/f'(x{r})
Select a reasonable guess for x0.
(Though personally I wouldn't use excal for this kind of things as I've heard it uses it's own weird fp mumber representations and algorithms, not IEEE ones.)
x = 2.156512353, this is from a solution using an old version of Derive.