Exponential equations e.g. $x^{x^2} = 8x$

Question

How does one go about solving an exponential equation with variable as base and in the exponent, such as:

$$x^{x^2} = 8x$$

I've seen people write (and am happy with): $$x^{x^2-1} = 2^3$$

Followed by: $$\Rightarrow x^2-1 = 3$$ $$\Rightarrow x = 2$$

This happens to be the right answer. But isn't the method wrong as the bases must be the same to justify this step?

Have I missed something? Is there another method to apply here?

Apologies for the straight forward question. I'm self-studying and you good people are the best for me to ask.

Answer 1

Assuming $$x \ne 0$$

$$x^{x^2} = 8x$$

$$\frac {x^{x^2}}{x} = 8$$

$$x^{x^2 - 1} = 8$$

$$x^{x^2 - 1} = 2^{2^2 - 1}$$

Now on comparing we get $$x = 2$$

Answer 2

Taking logs, we write $f(x) = (x^2-1) \log(x)- \log(8)$. Our problem has the solutions $f(x)=0$ which occur at $x=2$ and $x \simeq 0.121185$. Taking $f''(x) = 2\log(x) - (x^2 - 1)/x^2 + 4$ one observes that $f''(x) > 0$ for all positive $x$. So $f(x)$ is convex and there will be no other solutions $f(x)=0$.