A Nice Math Olympiad Exponential Equation 3^x = X^9
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- Опубликовано: 17 ноя 2024
- A Nice Exponential Equation 3^x = X^9 How to Solve Math Olympiad Question 3^x = X^9 Exponential Equation? What is the value of x? Find the value of x? Solving Math Olympiad Question A Nice Exponential Equation 3^x = X^9
In this video, we'll show you How to Solve Math Olympiad Question A Nice Exponential Equation 3^x = X^9 in a clear and concise way. Whether you're a student learning the basics or a professional looking to refresh your knowledge, this video is for you. By the end of this video, you'll have a solid understanding of how to solve math olympiad exponential equations and be able to apply these skills to a variety of problems. So whether you're studying for an exam or just looking to improve your math skills, be sure to watch this video and take your math knowledge to the next level!
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Nice, but...when we are talking about solving equation, it is necessary to find all of them or indicate that it's required find some of x. From graphs of functions easy to see that the equation has two solutions. One is : 1.15
Ok RS Grewal
I correct myself: the grade is nineth so the ecuation has 9 solutions)
Both functions have only one intersection point in the first quadrant. The solution shown is therefore completely correct.
@@bikox4352They have two intersection points. One being in x=27, and second in x≈1.151, which was mentioned
@@bikox4352yea, in the first quadrant maybe, what about the other ones?
Nice trick, but there's a flaw in the end when you deduce from x^(1/x) = 27^(1/27) that x=27. There's no warranty that it's the only solution, you'd have to prove that the function f(x) = x^(1/x) is injective for that. And it's not the case. It's the same mistake as when you have x² = 2², and you deduce from it that x = 2. You missed that x = -2 was also a solution.
Ok RD Sharma
Interesting
@Altair705 Both functions have only one intersection point in the first quadrant. The solution shown is therefore completely correct.
@@bikox4352 Sorry but no, as stated by others there are 2 solutions: x=27, and x≈1.150825. If you just drew the functions you probably missed the intersection point at x=27 because 3^27 is a huge value.
Even if there was a single solution what I said still applies: that method allows to find a solution but you have to justify that there's no other one.
Yeap, if you replace x for 27 on the third step you can tell there's something wrong already@@Altair705
The two real solutions using the Lambert W function are x = 27 and x = 1.150825 (6 decimal places). There are of course and infinite number of complex solutions as well.
You don’t use a real solution to the 6th decimal unless you are going to find all infinite possibilities
@@tobiasketteringham7201
Yes you do. You are obviously not an engineer!
@@davidbrisbane7206 and you are OBVIOUSLY not good at english, but regardless I would love to see how you came up with this basic question without a calculator and calculated it to the 6th decimal? As stated this is a basic university level question.
@@tobiasketteringham7201
There's a series you can use to compute the Lambert W function branches W(0) and W(-1).
By the way, I'd be interested to see how you calculate e^x without a calculator 🤣😂🤣😂.
@@davidbrisbane7206 You, pen and paper are turing complete. Everything your computer calculates you can too. e^x can be calculated using a series as well.
The funny thing is the note at the bottom saying you should know this … pride comes before the fall when you are dealing with math geeks and there can be multiple solutions.
It's easy to see x > 0. Now, notice the given condition is equivalent to x ln(3) = 9 ln(x) or ln(x) / x = ln(3) / 9 = ln(27) / 27. Because the first derivative of ln(x) / x is (1 - ln(x)) / x^2, we know ln(x) / x is increasing on (0, e) and decreasing on (e, infty). Thus, observing that x = 27 > e is a solution implies that there is one other solution in (0, e).
Can you please explain how you deduced ln(3)/9 = ln(27)/27?
@@eobardthawne6903you multiply fraction by 3/3 so you get (3 ln3)/27, then you put 3 in ln -> ln(3^3)/27, 3 cubed is 27.
@@MrKrabbs2009 now I get it. Thanks.
My first thought was using logarithms
@@Aleblanco1987 Same! That seemed like the only way that made sense to me before watching the video.
After watching the video though, I'm not disappointed with their method, but not satisfied either if you know what I mean, hehhe 😅
It was a nice & short way of solving it, but not a complete one
just cube each side and you get 3^(3x) = x^27. Then, (3^3)^x = x^27. Finally, 27^x =x^27 and then it's really easy to see x=27 is a solution. However, 27 isn't the only solution as another commenter already pointed out.
yeah, my initial thought was "there's gotta be an easier method with cubing", thanks for writing it out!
Agreed. Solving a few steps and doing what could've been done on the first step... This looks more like a hit and trial method rather than solving.
You lost me at x^27=27^x.
Lets say 27 is 3 and x is 2.
You're saying 2^3=3^2.
Therefore 8=9.
And how do you already know from that that x=27? Is it because 9 and 3 are the only non-hidden values and through pattern recognition x is obviously 27?
This math yt community is insane man
@@jshrekmaster no, obviously not - the statement is that through these transformations you would reach the statement x^27 = 27^x, which could only possibly hold true for one value of x, namely, 27. How you extrapolated what you extrapolated is genuinely beyond me
@@jshrekmaster x is an arbitrary variable that can be replaced with any number. 27 is a number that can't be replaced without any algebraic justification.
So, you can only replace x and not 27. when you set some random number to the power of 27 is equal to 27 to some random number, just looking at it 27 pops out as an easy solution since 27^27 visually is the same as 27^27.
The reason why I cubed each side is because 9 and 3 are coincidentally factors of 27, and that 3^3 is 27, so by cubing each side, you transform a 3^x to a 27^x and an x^9 into an x^27 through properties of exponents.
There is another solution using the Lambert W function in addition to infinitely many solutions using imaginary numbers, but those solutions are well beyond the scope of this video.
What I'm not happy about is that your method of solving required that you knew 27 would be the solution in the first place.
That is exactly the reason I hate this method, in the end he basically guess the solution
@@littlewizard9138 Yes I thought the same thing. I just kept yelling at the silent screen saying WTF?!!!
The only thing I can deduce from this and another video is you need to get a/a ratio. How dou you get this a?
In this case:
3^a = 9*a
In the other video they have 2 instead of 3 and 32 instead of 9.
So, they have 2^a = 32*a.
Being a = 8, one solution is 256.
Its important to note that 9 is power of 3, and 32 is power of 2.
Which is why I always hated calculus, pre calculus, calculus 2, physics, and math classes in general. College professors taught it as if you can and are supposed to see the solution and know it and then prove it. It’s why you won’t see me ever touching Math again unless it’s so important for something 😂
Ni puta idea
@1:32 you already know the answer that's why you multiplied & divide by 3.
That made me mad
Exactly
You are missing solutions. If you try x=1 and x=2 you can see that there is another solution between these values of x.
Wao
Yes, I graphed 3^x = x^9 on desmos, and it gave 1.15... and 27 as solutions.
doing this before the video
x^-9 * 3^x = 1
x^-9 * e^ln 3^x = 1
x^-9 * e^x ln 3 = 1
(x^-9)^-1/9 * (e^x ln 3)^-1/9 = 1^-1/9
x e^-(x ln 3/9) = 1 | you might have noticed i only used one of the roots. The W function actually handles the rest of these roots. im using the principal for now.
-(x ln 3/9) e^-(x ln 3/9) = -ln 3/9
Lambert W Function:
W(-(x ln 3/9) e^-(x ln 3/9)) = W(-(ln 3)/9)
-(x ln 3/9) = W(-(ln 3)/9) | W(xe^x) = x
x = (-9W(-(ln 3)/9))/ln 3
The Lambert function is funny. Afaik it's just the numerical calculated inverse of the definition.
So no closed form and actually cheating by claiming, the result was not estimated numerically 🤔
@@zyklos229 some people actually consider it a closed form because of its two representations of the principal:
W(x) = Σ [from n=1 to ∞] [x^n * (((-n)^n-1)/n!)]
and
W(z) = z/2π * (integral [-π to π] [((1 - v cot v)^2 + v^2)/z + (v csc v) e^-v cot v] dv)
but whether this is closed is debatable sooo yeah
@@zyklos229 The W function has nothing to do with numerical approximation. It is defined as the inverse of f(x)=x*e^x. Many functions are defined in this manner, as the inverse of another function. Logarithms, radicals, and inverse trig functions are examples.
Using Lambert W function , I got the following solution: x=exp^(-W(-Ln[3]/9))=> x=1.15082 which is not the only possible solution. But, it does satisfy the equation.
A better way to do this is to put the equation in Lambert form. For W0 I get approx. x = 1.151. For W-1(the minor root) I get x = 26.989. However, if you plot the graph, I don't see how 26.989 = x is a solution. If anyone on this thread can explain the contradiction, please jump in.
Why are you able to multiply both the radicant & index by 3? Just to get the equation to work?
Hmm... Here's another way: Assuming x is a positive integer, 3^x is a power of a single prime (3). By unique factorization in the integers, x^9 must also be a power of 3, and hence x must itself be a power of 3. So we can write x = 3^n for some positive integer n. Insert this in the original equation and we get 3^(3^n) = (3^n)^9 = 3^9n. By comparing exponents, we see that 3^n = 9n. It's easy to see that this only works for n = 3 (just draw and/or think: LHS is an exponential eq. while RHS is linear). Hence x = 3^3 = 27.
正确的解题思路和方法是对等号两边同时取Log, 将复杂的指数形式变成对数形式，经运算后即可得到解。
La fonction w de Lambert nous permet d'obtenir les deux solutions dont x=27
Could you tell me your pen's brand and also its model? Thanks
Pilot V7
Take x=3^t, now by comparing both sides you will get 3^t=9t, take t=3^y -eq(1) and now the equation becomes 3^t=3^(2+y), from which you get t=2+y -eq(2), now find intersection of eq1 and eq2 you will get that eq2 is a tangent of eq1 at y=1, then by putting y=1 you get t=3, which makes x=3^t=27 and this x satisfies our original equation, hence solved. 🎉
Let x = 3^t , so 3^t = 9t
Let t = 3^s, so 3^s = 2 + s → s = 1
So t = 3¹ = 3, then x = 3³ = 27
Correct, and I think this method is better than the one in the video. However, you are implicitly assuming that x is a positive integer (which the video also assumes), but there is another real positive solution.
I like this method of solving best! Substitution all the way!!!😂
This substitution is justified if you assume x is a positive integer. Just use unique prime factorization. From 3^t=9t you could have found 3^(t-2)=t, which has a single solution: t=1.
I am starting to feel that random things appeared just to prove LHS = RHS
You scored 50 out of 100.
This is why we learn logarithms & the natural logarithm (ln). Moreover, this is not at the level expected for an Olympiad. 😑
In Korea, power & exponential equations are covered in SOPHOMORE year of HIGH SCHOOL.
x = 3^(x-2). Let x =3^a (x>0 as 3^anything > 0). Thus, a = 3^(a-2). Let f(a) = 3^a/a. f is decreasing for a
Is this really a trick to solve the problem or just something that's straightforward once you already know the solution? It's like showing someone how to solve to a maze as - "just turn here then here then here!"
When he changed the 1 for 3/3 it was very clear he already knew the answer 🫠
practice automatically makes you know such techniques
Using properties of exponents in a quick fashion, I like it!
I would avoid posting incorrect math. next time don't assume injectivity.
It's not incorrect math though... next time tey thinking before writing a comment....
there's an unmentioned root at around x=1.16. the injectivity assumption in the math doesn't find this root, and instead only finds a single answer@@King1Z7
@@King1Z7 from x^(1/x) = 27^(1/27) that x=27 you'd have to prove that x^(1/x) is injective, which it isn't.
Damn thats really interesting, i'd be stuck with "x/log x = 9/ log 3" 😅but yeah as others said, you kinda assumed it's got only one solution.
Great approach….👍 Thank you very much much Sir
Elegant solution. I have an ugly solution with logarithms.
Log_3 of both sides
You get x = 9log_3(x) Call this *
Let log_3x = y, therefore x=3^y. Call this **
From * you have x = 9y
From ** plug in for x and get
3^y = 9y.
Compare this to the original equation. Right side no longer has exponent, so it's a bit better.
It's clear that y = 3 is a solution.
Plug this into equation x=9y and you get x=27
Other solutions? Don't know.
No, you need to use ln
Sorry, but you cannot just use as condition that base and exponents must be equal on both sides. It is not the only solution of the problem
Ok Ramanujan
The solution is logic
Both functions have only one intersection point in the first quadrant. The solution shown is therefore completely correct. The result of 3 power x is only positive for real numbers. The result of x power 9 is positive for positive real numbers and for negative real numbers is it negative and if x=0 is the result zero.
@@bikox4352bro is really spaming the same wrong answer under every comment lmao
@@bikox4352 x^(1/x) is not injective
Logarithm left the chat
You need to use lambert W function for another x
Cool video, please continue to make more
Brilliant approach. Well done 👌
You can use log to solve it quicker. Nice one. 😊
After seeing the solution, I am glad that I got the solution x = 27 but was not able to get the other solution.
Here is my solution. I found the solution of x = 27 by using simpler reductions until I was able to guess.
I knew that there was another solution because f(y) = 3^y and g(y) = 9y intersect at y=3 but f'(3) is irrational and g'(3)=9 so the line and the exponential function must intersect at some other point.
You can see from my work that I got stuck at the end since my reductions repeated and I was not able to get anything more.
"
3^x = x^9
x > 0.
x = 3^y
3^{3^y} = {3^y}^9=3^{9y}
3^y = 9y
x=9y
9y = x = 3^y
3^y = 9y
y>0
3^y = 9y, equal at y=3
y = 3^z
3^(3^z) = 9*3^z
3^(3^z) = 3^{z+2}
3^z = z+2
z>-2
z=1->y=3->x=3^3
3^(3^3) ?=? (3^3)^9
3^(27) ?=? 3^{27} YES
one more solution
3^z = z+2
z is odd,
z = a-2, a
eq 3
z > -2 => a > 0
3^{a-2} = a
3^{a} = 9a
same equation as for y
Try to solve
a = y
a = 3^z
a = 3^{a-2}
9a = 3^a, same equation as for a.
So, a = y
"
Thank you sir 🙏
Most questions usually accept the multiplication of numbers that are in the question as the answer. In this case 3*9, there can be many answers but an easy one
Nice trick ❤
Can We can take log both side and solve ?
First off where do all of these other numbers come from? It’s like they came out of the blue out of thin air. Also, what would be the best place for me to learn math wether it be a website or a book or course etc.
There is another solution , i calulated it numerically, about x=1.14 can you show the calculation? Thank you
Can we take log to solve easily if calculater is allowed.
If you replace the x with any number they don't match so does that make it unusable?
What pen are u using???
Seeing the X’s being written like that physically pained me.
"Physically"
@@unarmed_civilian I felt it in the deepest depths of my inners.
Sounds like pedantism to me.
@@haarp9069 not at all. It’s improper mathematical notation
@@him050 Definitely pedantism.
Hi, thanks for the clever trick, but ... As many people already said, it's not the only solution : ~1.1508 is also a valid solution.
Also, at 1:34 how are we supposed to know when to do that? It's very interesting but I'm never gonna know when and how to use that 😢.
PS : thank you anyway
Wonderful......!🎉🎉
my try：
make a^3ln3=9aln3
😊😊
x^( 1/x) = (3 ^ n) ^ ( 1/ 9 n)
Any feasible solution to the same satisfy 3 ^ n = (3 ^ 2) *n
3 ^ ( n - 2) = n
n = 3 has a trivial solution to the same
Hereby x = 3 ^ 3 = 27
I guessed and got 27.
Can even do the same sum by logarithms right?
Nice Video Please continue uploading new videos
This "tricj" does work for 6^x=x^7 ?
Or we should know the solution before ?
Just do the formula you learned in your school
@@BananaaChan can you show me how you do to solve this one ?
@@Frank-kx4hc To find a solution to the equation 6^X = X^7 we can use the trial and error method or use other numerical methods such as the Newton-Raphson method. However, let's use a trial and error approach to find a solution numerically:
1. Choose an initial value for X, for example X = 1.
2. Calculate 6^X and X^7:
6^1 = 6
1^7 = 1
Since 6 is greater than 1, we know that the value of X needs to be increased.
3. Select a new (X) value, for example X = 2.
4. Calculate 6^X and X^7:
6^2 = 36
2^7 = 128
Since 36 is still smaller than 128, we know that the value of X needs to be increased again.
5. Repeat steps 3 and 4 with the new X value. Continue this process until we get close to a solution.
After several iterations, we will find that X ≈ 6 satisfies the equation 6^X = X^7
@@BananaaChan thank for your answer.
Of course, Iteration method leads all time to an approximative solution.
Here, there is a solution between 1 and 2 bcse 6^1 > 1^7 AND 6^2 < 2^7.
However the "trick" of video is unable to find this soution.
So such "trick" works for a solution that we KNOW BEFORE !!!
Notice that an APPROXIMATIVE solution is not a TRUE solution.
@@Frank-kx4hc Sorry if I'm wrong, because I'm still in junior high school and I have to study hard, especially since I can't speak English well. I typed this message with translation from google. And, I'm still learning about algebra, not approximation.
Isn't there a more, how do I put it, layman solution. Since we are talking about exponentiation, it's evident that x is a power of 3. Just input values and reduce to check if it fits. Put x as 3, 9, 27 and so on. Try lower powers as well.
一応x≠0であることは示したほうが良いんでしょうか？
Nice i need this trick❤
Thanks for sharing!
More easy method would be:
3^x = x^9
x ln3 = 9 lnx
Now just put x = 3, 3^2, 3^3, ... until equation satisfies
You have a logical fallacy when you say “you should know this”. Most of the regular folks clicking on this video don’t know ‘this’. 😊
Is it possible to state the following?:
If A^B = B^C then B=A*C
I'm not at all math inclined but I just bappen to notice 27 is 3*9 and they also happen to be in the base question
This is an interesting observation. The next logical step is to see if you can find a counterexample, that is, can you pick values for A, B, C such that the first statement is true but the second one isn't?
In this case, if you choose A=2, B=2, C=2, you can see that 2^2 = 2^2, but it is not the case that 2 = 2 × 2.
So in this case your assertion does not work. But I think it is great that you are observing patterns, because that's what people who are "math inclined" do.
Replacing x with 3^n, I was able to reduce it to 9n=3^n, but then I was stuck (even though I could see that 3 was a solution).
Nice trick with adding 3/3👍
Easy way to solve can be
Taking log both side with base 3
ln*
Nice job
I'ts the same to use one solution in equations of 2° degree when b^2-4ac#0
Quite impressive
Which to me seems exactly equivalent to solving it by simply trial and error substitution of integers. Am I wrong?
Amazing！！！！！！！！！！！！！！！！！！！！！！！！！！！
Lovely question
Lemme focus
*Focused*
*Focused*
*Focused*
The value of x is 3
We can take log on both sides and then multiply 3/3.
Who solved it in the thumbnail itself, I hardly took 20 sec
Thanks
Why conveniently represent the 1 (in step 3) as 3 by 3, why not 9 by 9 that way x=1. And since when can you carry out operations on only one side of an equation
Very Nice
Myslím vhodné pro použití Lambert W funkce.
It is very easy😅
When calculating fraction multiplication, don't the denominators have to be the same?
Nope, only fraction addition and subtraction.
I have another easy method to slove it ।
3^x= X^9
We can write in this form
3^1/9 = x^1/x
After that we can follow the methods which mention in the video but we can save more step
Just when I thought I escaped advanced math it shows up in my recommended. 😂
Do I have to do it left handed?
Much easier to use logic. If we stay w whole numbers, X has to be root 3 and larger than 9. First up is 27. Falls into place.
I have been thinking about this for a long time, for me the second solution of approximately 1.15 makes perfect sense. But 27? I really can't understand how 3^27 could equal 27^9. Every time x gets bigger it 3^x and x^9 grow further apart. I've tried putting them in a graph, sadly the shear size of the numbers makes it hard to read.
Just consider expanding the exponents. 3^27 is 3 times itself 27 times. 27^9 is 3*3*3 times itself nine times. Either way, you have 27 threes in a row.
Now I can sleep better knowing this is solved.
How do we know there are no more solutions?
We don’t. In fact, there is one other real solution.
X=9•3
😎
I solved this in my head…I’m very intuitive…in a strange way the answer just came to me then I checked it and it worked
I’m a genius
So does your “genius” give you the other solution that not easy to found?
I didn’t approach it as a math Olympiad scenario where my solution will be graded
I approached it as can I find a solution by just looking at the problem and thinking about it subconsciously and not actively “solving it”...it was more of a way to test my intuition...it worked by that standard
@@littlewizard9138fair question btw
Very nice
Wow! Never seen this before!
Fantastic
What song did you use for this video?
Thank you...clever
I found the way of calculating x very confusing and I solved it, I think logically but please let me know if my reasoning isn't sound. The answer to 3 to any power must only have 3 as a factor - and the answer to LHS can only be 3. 9. 27, 81 etc for the RHS to only give an answer with only 3 as a factor, x must itself be only a multiple of 3. and in fact, so as not to incorporate any otherf factors in the answer must only be a multiple of 3's - so 3,9,27 etc. X obviously cannot equal 3 (3 to power 3 is not equal to 3 to power 9) and also obviously not 9 as 3 to power 9 cannot equal 9 to power 9. 27 is the next logical answer to try and since 27 gives 3 to the power 3, then 27 to the power 3 can be written as 3 to power (3 x 9) which is s to power 27 on both sides. It took me longer to type this out than to do it in my head. Can't understand how this video has been presented this way unless I am missing something?
Hi, what do you have a pen ?
can you please do this :
(5x)^x=5^3125
x^2=3125 => x= + - sqr 3125
@@duongtrannam3127 yoo i solved it its 625 but ty
Carry on ........
We can take log3 that is log with base 3 both side and try to solve it, finaly we will get X/log3X = 9 and then finally we will get to know x=27
Excelente. Incrível.
How do you know that the function x^1/X is one to one (bijection function)
It is not
Jesus I felt the fear of maths again after twenty five years 😂