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.

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.
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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 $\log_{\frac 1b} (x) = \log_{\frac 1b} (b) \log_b (x) = -\log_b x$.
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[1](Which is how you prove the change of base theorem:  $\log_n k = \frac {\log_m k}{\log_m n}$).
