How to show that $- \log_b x = \log_{\frac{1}{b}} x$ I saw the following log rule and have been struggling to show it's true, using the change of base rule. Any hints for proving it would be much appreciated. 
$- \log_b x = \log_{\frac{1}{b}} x$
I get as far as showing that $- \log_b x = \log_b \frac{1}{x}$ and think the change of base rule for logs, $\log_b x = \frac{\log_c x}{\log_c b} $ might be useful, but am not sure how to proceed. 

Back to basics:
Let $y:= \log_{1/b}x ;$
$(1/b)^y = (1/b)^{\log_{1/b}x} = x;$
$\log_b(1/b)^y = \log_b(x);$
$y\log_b(1/b) =y\log_b (1) - y\log_b (b)$
$= -y = \log_b(x).$
Compare with first line:
$y= \log_{1/b}x = - \log_b(x).$
