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Find all solutions for \(x\) in \( 2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 . \) Prove your answer

 May 31, 2019
edited by Guest  Jun 1, 2019

I'm pretty sure that you mistyped that; the answer is unknown.

 Jun 1, 2019
edited by CuriousDude  Jun 1, 2019

I checked and there are no mistypes.  I have changed it to latex so you can look at it easier.

Guest Jun 1, 2019

Oh, the \(LaTeX\) version is much clearer.

Simply both sides first so you can isolate the variable.

Remember, isolate, isolate, isolate.


\(x\) can be -1, 0, or 1.smiley

CuriousDude  Jun 2, 2019

The answer is correct but is it possible that you could go into further detail on how you ended up with your answers?

Guest Jun 2, 2019


\(2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 \\ 2(2^x- 1) x^2 + (2^{x^2}-2)x = 2(2^{x} -1) \\ 2(2^x- 1) x^2- 2(2^{x} -1) =- (2^{x^2}-2)x \\ 2(2^x- 1) (x^2-1) =- (2^{x^2}-2)x \\\)

By inspection,

If both sides equal zero then they both must be the same.

If x=0 both sides are zero so that is a solution.

If x= +1 or -1 they are solutions too.


If x is bigger than 1 then the left side is positive and the right side is negative so no answers there

If x is less than -1 then the left side is negative and the right side is positive so no answer there either.


So could there be any more answers between -1 and +1?

I do not think so but i do not know how to prove it.


Maybe CuriousDude can show us both.    laugh

 Jun 7, 2019

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