Solving $x^{\log_{10}(x^2)} = 100$ $x^{\log(x^2)}=100$
Edit: I am writing $\log$ for $\log_{10}$.
I tried to solve this and can get the solutions $10$ and $10^{-1}$. But how can I get the solutions $-10$ and $-10^{-1}$? And other solutions (including complex), if exist? I'm not sure how to use the complex logarithm in this problem.
My attempt:
Note that $x \neq 0$, because we have in the equation $\log(x^2)$.
Assume $x &gt; 0$. Then:
$x^{\log(x^2)} &gt; 0$
$\log (x^{\log(x^2)}) = 2$
$\log(x^2)\log(x)=2$
$2\log(x)\log(x)=2$
$\log^2(x) = 1$
$\log(x) = \pm 1 \implies x=10$ or $x = 10^{-1}$.
But how about $x&lt;0$?
For real $x,$
$$x^{\log_{10}(x^2)}=x^{2\log_{10}|x|}=(|x|^2)^{\log_{10}|x|}$$
Let $\log_{10}|x|=y\implies|x|=10^y$
So, we have $$10^2=10^{2y^2}\implies y=\pm1, |x|=10^{\pm1}, x=\pm(10^{\pm1})$$
\begin{align}
x^{\log x^2}&amp;=100\\
2(\log x)^2&amp;=2\\
(\log x)^2&amp;=1\\
\log x&amp;=\pm1\\
x&amp;=10^{\pm1}
\end{align}
As this is a quadratic equation in $\log x$, given the monotonic nature of the logarithmic function, we can conclude that $10^{\pm1}$ are the only solutions.