<s>[INST] Find all $x$ such that $2^x,2^{x^2}$ and $2^{x^3}$ form $3$ terms of an A.P. 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.
 [/INST] 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.
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