How to solve this equation ? - Exponential equation

Question

How do i solve that equation algebraically ?

Thanks!

Answer 1

IN THE CASE $x\in\mathbb{R}$

Firstly you can factor:

$2^x(x^2+2^x)=(x^2-2^x)(x^2+2^x)$, but $x^2+2^x\neq0$ so we simplify:

$2^x=x^2-2^x$ and we obtain just: $2^{x+1}=x^2$.

So you have to find zeroes of $f(x)=2^{x+1}-x^2$ with $x\in \mathbb{R}$. We have that $x=-1$ is a solution.

$f'(x)=\log(2)2^{x+1}-2x$, which is always positive (clearly it's positive for $x<0$), and it's not difficult to verify for $x>0$ So $f$ is monotonic and $-1$ is the only solution.