How do I prove that $\log(1/b)^x= -\log b^x$? How do I prove that $\log(1/b)^x= -\log b^x$?
I am working on logarithm properties. And I have come across one of the power rules, where the base is a fraction. I'm struggling to make assumptions and prove that the two equations are equal. Someone to help.
 A: We can use the following laws of exponents  $$a^{-m}=\frac1{a^m}, \qquad (a^m)^n=a^{mn}=(a^n)^m$$
and the power law of logarithms $$\log_b x^n=n\log_bx.$$
Solution:

We have $$\log\left(\frac1b\right)^x=\log(b^{-1})^x=\log(b^x)^{-1}=-\log b^x$$ as desired.

A: Hint:
For any $\;a,b\in (0,\infty)\;$ , we have that
$$\log\frac ab=\log a-\log b$$
But we also know that $\;\log 1=\ldots?$
A: 
You want to show:
$$\log(1/b)^x= -\log b^x$$

We need to use two Theorems for this.
$\bullet~$Theorem 1: Let $a, b > 0$. Then $\log(a/b) = \log(a) - \log(b)$
Proof: Let, $u = \log a$ and $v = \log b$. Let the base be $10$ wlog. Therefore from the definition of $\log$, we have that
$$ \log a = u \implies 10^u = a \quad \text{ and } \quad \log b = v \implies 10^v = b $$
Hence, $$ \frac{10^u}{10^v} = 10^{u - v} = \frac{a}{b} \implies u - v = \log\left(\frac{a}{b} \right) = \log a - \log b $$
$\bullet~$Theorem 2: Let $a > 0$ and $m \in \mathbb{R}$. Then $\log(a^m) = m\log(a)$
Proof: Let $u = \log (a^m)$ then we have that
$$ u = \log (a^m) \implies 10^u = a^m \implies 10^{u / m} = a \implies \frac{u}{m} = \log (a) \implies u = \log(a^m) = m \log (a) $$

Thus by Theorem 1 and Theorem 2 we have $$\log(1/b)^x= x (\log(1) - \log(b) ) = -x \log (b) =  -\log b^x$$

Edit: $\log 1 = 0$.
proof: Let's take $u > 0$. Then
$$ u^0 = 1 \implies 0 \cdot \log(u) = \log(1) \implies \log(1) = 0 $$
A: Claim:  $\log_{\frac 1b} x = -\log_b x$.
Pf:  $b^k = x \iff (\frac 1b)^{-k} =x$.
So $\log_b x = k \iff \log_{\frac 1b} x = -k$.
That's all.
=====
Alternatively the change of base theorem stats $\log_m k = \frac {\log_b k}{\log_b m}$ so $\log_{\frac 1b} x = \frac {\log_b k}{\log_b \frac 1b}=\frac {\log_b k}{-1} =-\log_b k$.
Of course we have to prove the change of base theorem.
Theres also this:  $\log_m (n)\log_n k = \log_n (k) \log_m n = \log_m n^{\log_n k} = \log_m k$[1].
So $\log_{\frac 1b} (x) = \log_{\frac 1b} (b) \log_b (x) = -\log_b x$.
=======
[1](Which is how you prove the change of base theorem:  $\log_n k = \frac {\log_m k}{\log_m n}$).
