unclear step in a logarithmic expression simplification routine (trivial question)

Question

I don't understand why the value there is $\frac14$ and not $-\frac12$. The base of the second logarithm could be viewed as $a$ with power equaling $-1$, couldn't it? Then its multiplication by $\frac12$ should give $-\frac12$.

From my textbook's "solution volume" (the task is to simplify the expression):

Answer 1

We have $$log_{1/a}^2\sqrt{a^2-1}=(log_{1/a}\sqrt{a^2-1})^2=\left(\frac{log_{a}\sqrt{a^2-1}}{log_{a}{(a^{-1})}}\right)^2=\left(\frac{\frac12log_{a}{(a^2-1)}}{-log_{a}{a}}\right)^2=\frac14\left(log_{a}{(a^2-1)}\right)^2=\frac14log_{a}^2{(a^2-1)}$$

Answer 2

Hint:

Use the change of base formula: $$ \log_b(x)=\dfrac{\log_a (x)}{\log_a (b)} $$

In your case (as an example) $$ \log_{\sqrt[3]{a}}(x)=\dfrac{\log_a (x)}{\log_a (\sqrt[3]{a})}=3\log_a (x) $$