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?
 A: Let $x>0$, then $\log_b(x)$ is the unique real number such that $b^{\log_b(x)}=x$, but $b^{-\log_{1/b}(x)}=\frac{1}{b^{\log_{1/b}(x)}}=\left(\frac{1}{b}\right)^{\log_{1/b}(x)}=x$, therefore $-\log_{1/b}(x)=\log_b(x)$.
A: You've written your formulas in a way that is confusing you.  You wrote  $$\log_{1/2}(4)=\frac{1}{2}^y=4$$ but what you actually meant was $$\log_{1/2}(4)=y \quad\text{ because }\quad \frac{1}{2}^y=4.$$
Then for the other version you have:
$$-\log_2(4)=-2^y=4$$
but it should be:
$$-\log_2(4)=y\quad\text{ because }\quad ???$$
Let's move that minus sign:
$$\log_2(4)=-y\quad\text{ because }\quad???$$
Okay, now it looks like the first one, and we get:
$$\log_2(4)=-y\quad\text{ because }\quad2^{-y} = 4.$$
And $2^{-y} = 2^{-1\cdot y} = \left(2^{-1}\right)^y = \left(\frac12\right)^y$.
I'm not sure that gives any intuition, but at least it shows where your mistake was.
A: We will use the following two identity
$$\log_a(b)=\log_{a^x}b^x$$
$$log_a(b^n)=n \log_a(b)$$
We will start with the question
$$log_b(x)=-log_{\frac 1 b}(x)$$
Seeing $\frac{1}{b}$ as the base in the right hand side of the equation motivates us to use the second identity setting up $x = -1$.
$$log_b(x)=-log_{\frac 1 b}(x)=-log_b(\frac{1}{x})$$
Now we get rid of the minus sign using the second identity
$$log_b(x)=-log_b(\frac{1}{x})=log_b((\frac 1 x)^{-1})=log_b(x)$$
A: This comes from the following logarithmic identities:
$$\log(\frac{x}{y}) = \log(x)-\log(y)$$
$$\log_b(a)=\frac{\log_x(a)}{\log_x(b)}$$
where $x$ is any positive number (except 1).
Then you can calculate
$$\log_{1/b}(x) = \frac{\log_b(x)}{\log_b(1/b)} = \frac{\log_b(x)}{\log_b(1) - \log_b(b)}$$
Now remember that $\log_b(1)=0$ for any allowed base and $\log_b(b)=1$. So this simplifies to
$$\frac{\log_b(x)}{0 - 1} = -\log_b(x)$$
A: $$log _{\frac 1 b} (x)=log _{\frac 1 b} (x)$$
$$x=(\frac 1b)^{log _{\frac 1 b} (x)}=b^{-log _{\frac 1 b} (x)}$$
We take log in base $\frac 1{\frac 1b}=b$ of both sides, we get:
$$log_b (x) =-log _{\frac 1 b} (x) log_b b=-log _{\frac 1 b} (x) $$
A: 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}
