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$
 A: Hint: $\log_{\frac {1} {\sqrt b} }\sqrt x= -\log_\sqrt b \sqrt x=-\frac{\ln \sqrt x}{\ln \sqrt b}$
A: You got stuck because you made a mistake: $\left(\frac {1} {\sqrt b} \right)^n \ne \frac {n} { (\sqrt b)^n}$
Your proof should go like this:
$n=\log_{\frac {1} {\sqrt b} }\sqrt x$
$\iff \left(\frac {1} {\sqrt b} \right)^n = \sqrt x $
$\iff \frac {1} { (\sqrt b)^n}= \sqrt x$
$\iff 1 = (\sqrt b)^n \sqrt x$
Can you finish your proof now?
Edit:
I see you finished your proof, but you did something more complicated.
Your proof was almost done like this.
$\iff 1 = (\sqrt b)^n \sqrt x$
$\iff 1 = b^n x$
$\iff b^n = x^{-1}$
$\iff n = \log_b(x^{-1}) = -\log_b(x)$
A: \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}
A: There is only one logarithm rule you need to know, and that is the definition of logarithm:
$$\log_b x = y \iff b^y = x. \tag{1}$$
In other words, the base-$b$ real-valued logarithm of a positive number $x$ is an exponent $y$, to which the base $b$ is raised to yield $x$.  So for instance,
$$\log_3 81 = 4 \iff 3^4 = 81.$$
So if $$\log_{1/\sqrt{b}} \sqrt{x} = y, \tag{2}$$ then $$\left(1/\sqrt{b}\right)^y = \sqrt{x}.$$  But this is the same as writing
$$(b^{-1/2})^y = x^{1/2},$$
or $$b^{-y} = x.$$
Now apply the definition of logarithm again:
$$b^{-y} = x \iff \log_b x = -y. \tag{3}$$
Now we combine Equations $(2)$ and $(3)$:
$$\log_{1/\sqrt{b}} \sqrt{x} = y = -\log_b x,$$
and we are done.
While it may be convenient to use other formulas like the "change of base" formula or $\log_b x^m = m \log_b x$, in actuality these other formulas all follow from the original definition $(1)$.
