Find all $x$ such that $2^x,2^{x^2}$ and $2^{x^3}$ form $3$ terms of an A.P.

Question

I know that if $a,b,c$ are in Arithmetic Progression, then $2b=a+c$, but in this case, I am unable to solve for $x$. Hints are appreciated. Thanks.

Answer 1

HINT:

$$2^x(1+2^{x^3-x})=2^{x^2+1}\iff 1+2^{x^3-x}=2^{x^2-x+1}$$

If $x^3-x>0,1+2^{x^3-x}>1$ is odd unlike $2^{x^2-x+1}$ as $x^2-x+1>0$ for real $x$

So, $x^3-x$ must be $0$

and consequently, $2=2^{x^2-x+1}\implies 1=x^2-x+1$

Answer 2

Since $2b=a+c$ we have $2. 2^{x^{2}}=2^{x}+2^{x^{3}}$

which implies that $2^{x^{2}+1}=2^{x}+2^{x^{3}}$

Now taking log on both sides we get $x^{2}+1$ log2=log $(2^{x}+2^{x^{3}})$ solving this we can find $x$

Answer 3

For $x>0\land x\ne1$, $$0<x(x-1)^2\implies x^2<\frac{x+x^3}2.$$

Then by monotonicity and convexity of the exponential

$$2^{x^2}<2^{\frac{x+x^3}2}<\frac12\left(2^x+2^{x^3}\right).$$

For $x<0$, $$\dfrac12\left(2^x+2^{x^3}\right)<1<2^{x^2}.$$

Two cases remain, $x=0$ and $x=1$, that happen to be solutions.