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 $x > 0$. Then:
$x^{\log(x^2)} > 0$
$\log (x^{\log(x^2)}) = 2$
$\log(x^2)\log(x)=2$
$2\log(x)\log(x)=2$
$\log^2(x) = 1$
$\log(x) = \pm 1 \implies x=10$ or $x = 10^{-1}$.
But how about $x<0$?
 A: 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 $$10^2=10^{2y^2}\implies y=\pm1, |x|=10^{\pm1}, x=\pm(10^{\pm1})$$
A: If $x_0$ is a solution to $$x^{\log_{10}x^2}=100$$ $-x_0$ is also a solution iff $\log_{10}x^2$ is an even integer.
Is  $\log_{10}x^2$ an even integer for the obtained solutions, $10^{\pm 1}$?
A: The problem stems from 2 things

*

*The fact that $\log$ is not defined when the the argument is negative

*We cannot work with negative powers raised to exponents which are variables which vary continuously ie $(-1)^x$ has a graph which oscillates indefinitely and hence we cannot work with it.

The moment you take $2$ out from the exponent of $x^2$, you lose the negative solutions. The same thing happens when you take $\log$ on both sides. I don't think there is any way to solve this for those values. I think the best bet is to just check and verify for the inverse counterparts once you get the positive solutions. And this problem cannot be resolved by taking complex numbers into account, this is a problem due to the ill-defined $(-1)^x$. If there are complex solutions, you can solve for them by replacing $x$ as $x+iy$ but you will surely miss solutions like these. Hope this helps.
A: $$x^{\log(x^2)} = 100$$
$$\implies \log(x^2) = \log_x{10^2}$$
$$\implies \log(x^2) = 2\log_x{10}$$
$$\implies \log(x^2) = 4\log_{x^2}10$$
$$\implies \log(x^2) = \dfrac{4}{\log(x^2)}$$
$$\implies \log^2(x^2) = 4$$
$$\implies \log(x^2) = \pm 2$$
$$\implies x^2 = 10^{\pm 2}$$
Now you got the $4$ roots: $\pm 10; \pm \dfrac{1}{10}$
A: This may help :
$$x^{\log_{10} x^2}=100$$
$$\log_{10} x^{\log_{10} x^2}=2$$
$$2\log_{10}^2 x=2$$
$$\log_{10} x=\pm 1$$
$$|x|=10^{\pm 1}$$
$$x=\pm 10,\pm \frac{1}{10}$$
A: \begin{align}
x^{\log x^2}&=100\\
2(\log x)^2&=2\\
(\log x)^2&=1\\
\log x&=\pm1\\
x&=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.
