Why $\log_b(x)$ = $-\log_{1/b}(x)$? IN a textbook I am asked to consider the graphs of $f(x) = \log_{1/2}(x)$ and $-\log_2(x)$. They appear identical and then the textbooks tells the rule that $\log_b(x)$ = $-\log_{1/b}(x)$.
I tried to prove this to myself on paper.
I can write both equations in exponential form. Pretend $x=4$:
$$\log_{1/2}(4)=\frac{1}{2}^y=4$$
Then write the same for the other version:
$$-\log_2(4)=-2^y=4$$
If I know these are equivalent then the two exponent versions should be the same no?
$$4=\frac{1}{2}^y=-2^y$$
It's not clicking and I don't 'get it' from here. Is my last equation true? Why are they equivalent?

Let $y=-\log_{1/b}(x)$. We see that
\begin{align}
&y=\log_{1/b}(x^{-1}) \\[4pt]
\implies&\left(\frac{1}{b}\right)^y=\frac{1}{x} \\[4pt]
\implies& \frac{1}{b^y}=\frac{1}{x} \\[4pt]
\implies& x=b^y \\[4pt]
\implies& y=\log_b(x) \, .
\end{align}
