<s>[INST] Vanier College Practice test - Prove that $\log_{\frac {1} {\sqrt b} }\sqrt x= -\log_b x$ Source : Colorado.edu, Vanier College, Worksheet Logarithm Function, Question 8 (https://math.colorado.edu/math1300/resources/Exercises_LogarithmicFunction.pdf)
To be proved :
$\log_{\frac {1} {\sqrt b} }\sqrt x= -\log_b x$.
What I did, before getting stuck :
$n=\log_{\frac {1} {\sqrt b} }\sqrt x$
$\iff \bigg (\frac {1} {\sqrt b} \bigg)^n = \sqrt x $
$\iff \frac {n} { (\sqrt b)^n}= \sqrt x$
$\iff n = (\sqrt b)^n \sqrt x$
Whith the help of @jjagmath I continue the proof at the point I made a mistake :
$ \frac {1} {(\sqrt b)^n}= \sqrt x $
$ \iff 1 = \sqrt x (\sqrt b)^n$
$\iff \frac {1} {\sqrt x} = (\sqrt b)^n$
$ \iff x^{-1/2}= (\sqrt b)^n$
$ \iff n = \log_{\sqrt b} x^{- \frac {1} {2} }$
$ \iff n = - \frac {1} {2}\log_{\sqrt b} x$
$\iff n =  - \frac {1}{2} \frac {\log_b x} {\log_b b^{1/2}}$ $\space \space \space$  by theChange of base formula.
$\iff n =  - \frac {1}{2} \frac {\log_b x} {\frac {1} {2}}$
$ \iff n = - \log_b x$
 [/INST] \begin{align}
\log_{\frac{1}{\sqrt{b}}}{\sqrt{x}} &= \frac{\log{\sqrt{x}}}{\log{\frac{1}{\sqrt{b}}}}\\
&= \frac{\log{\sqrt{x}}}{-\log{{\sqrt{b}}}} \\
&= \frac{\frac{1}{2}\log{{x}}}{-\frac{1}{2}\log{{{b}}}} \\
&= -\frac{\log{x}}{\log{b}} \\
&= -\log_b{x}
\end{align}
In total we use the following two rules:
\begin{align}
\log_ab &= \frac{\log{a}}{\log{b}}, \\
\log{(a^{c})} &= c\log{a}.
\end{align}
</s>