Show that:
$\displaystyle \log_{x^2}(y)$ = $\displaystyle \log_x\sqrt{y} \$
$\displaystyle x > 0$ and $\displaystyle y > 1$
Use the base change formula:
$\displaystyle \log_a(x)=\dfrac{\ln(x)}{\ln(a)}$
With your example you have:
$\displaystyle \begin{aligned}\log_{x^2}(y)&= \\
&= \dfrac{\ln(y)}{\ln(x^2)} \\
&=\dfrac{\ln(y)}{2\ln(x)} \\
&=\dfrac{\frac12 \ln(y)}{\ln(x)} = \dfrac{\ln(\sqrt{y})}{\ln(x)} = \log_x(\sqrt{y})\end{aligned}$
1. Please do yourself and do us a favour and start a new thread if you have a new question.
2. This one looks very ugly to me. Are you sure you typed it correctly here? (I'm only asking because the key "e" is exactly under the key"3" and maybe you slipped a little bit?) If the e is meant to be a 3 then your question becomes:
$\displaystyle 3^{x^2} = \frac{1}{\sqrt {3^{x^5}}} ~\implies~ 3^{x^2} \cdot \sqrt {3^{x^5}}= 1~\implies~3^{x^2+\frac52 x^2} = 1=3^0~\implies~\frac72x^2=0~\implies~x=0$