Math Olympiad Problem | Find X | X^2^X=X ^16| challenging Algebra Problem | Olympiad Mathematics

In this video, we will solve this algebraic equation problem trending 2022. The problem is find the value of x in X^2^X=X ^16. Math Olympiad problems
we will solve for all possible values of X. This is actually one of the challenging math problems so do not ignore this video.

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[philipkudrna]

This shall be an Olympiad question? I did it in 3 seconds: as the base is the same, you can equate 2^x=16, which leads to x=4. Of course x could also be zero (zero power smthg is always zero) or x could also be one (1 power whatever remains 1), but that‘s even more trivial!

[Nikioko]

16 = 2⁴ ⇒ x = 4.
Apart from the usual solutions 0^x = 0, 1^x = 1 and (-1)²ⁿ = 1, of course.
Solved in 1 second.

[suklaandanusartgallery]

This sum has a simple solution
X^2^X =X^16
=> 2^X = 16. [ as the base are same in both side ]
=> 2^X = 2^4
=> X=4 as the base are same in both side ]

[rorydaulton]

One quibble: at time 1:43 you change ln(x^(2^x-16)) to (2^x-16)*ln(x). That works only if you know that x is positive. But you have not established that. With a small change in your problem, x=-1 would also be a solution. (I know it is not a solution in your given problem.)

In general, given the equation a^b = a^c and using only real numbers, there are 4 cases: b=c; a=1; a=0 (with b>0 and c>0); and a=-1 with b and c rational numbers whose numerators are both even or both odd. Don't forget the possibility that the base is -1. Of course, enlarging the number system to the complex numbers adds additional cases.

In your particular problem, x=-1 does not work, so your three solutions are the other three cases.

[zainabhusain]

This is finally solved with so much integrity

[paull]

It’s x=4. If a number raised to a power equals the same number raised to a power, you can set the powers equal. 2^x=16. You can multiply the number 2 bu itself 4 times to get 16. So the answer is 4. Not a very difficult question.

[mathisnotforthefaintofheart]

I saw x=4 by inspection but try to graph this solution with any graphing device.... But there are three solutions, not one as some so called smarties are indicating in comments below

[gregsmith]

(X^2)^x = (x^2)^8. Therefore x=8 is also a Solution. Plug this value into original equation and it is correct.

[anroljames]

I got X= 8 because X^2^X = X^16. Can be written as X^2^X = X^2^8 by utilizing the law of exponents. Therefore X = 8. I did not see that answer expressed in any of the answers.

[gregsmith]

This is a worry to me. Using my method I only got one answer of x=8. On the video there were 3 answers, all 4 of which were correct. Is there a method of doing this equation and getting all 4 answers? If not, how do we know when we solve an equation that we have got all the solutions?

[ghgh-xfnu]

I think the only solution is only 4 not 1 or 0 because
1^n=1^16 and 0^n=0^16 and n have any value not only 1 or 0
If x=4 so 4^2^4=4^16 so 4^16=4^16
And the solution is X^2^X=X^16 because the base is equaled in two side so 2^x=16 log2^x=log16 log2^x=log2^4 xlog2=4log2 we remove log2 from two side so x=4 and this is the accurate solution in my opinion because is equaled in base and exponent

[touhami]

X=0 is not solution !!! Bcse 0^0 undefined.

[leoall]

No that wrong because
xsquare on x means
Put any nubmer give us 16
We suppose the x is 8
(x^2)^8 be x^16 = x^16 that correct

[Nikioko]

You can use the log to any base. In this case, log₂ would be good.