how to find number of solutions of $\frac{x^2}{8}=2^{3x}$

Question

how to find number of solutions of $\frac{x^2}{8}=2^{3x}$

this equation I can write as $x^2=2^{3(x+1)}$

I got one solution for this x=-1

How to find all solutions?

Answer 1

For $x\ge0$ there is no solution. For $x<0$ , $\dfrac{x^2}{8}$ is strictly decreasing and $8^x$ is strictly increasing so it has the only root you found.

For showing that $\dfrac{x^2}{8}<8^x$ we use the following lemma:

Lemma: for two functions $f(x)$ and $g(x)$ if $f(x_0)>g(x_0)$ for some $x_0$ and $f'(x)>g'(x)$ for $x\ge x_0$ then $f(x)>g(x)$ for $x>0$.

Using this lemma 2 times we conclude what we want:$$\forall x>0\to8^x(\ln 8)^2>\dfrac{1}{4}\to 8^x\ln 8>\dfrac{x}{4}\to 8^x>\dfrac{x^2}{8}$$