How to show that $- \log_b x = \log_{\frac{1}{b}} x$

Question

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.

Answer 1

we get $$\log_ \frac{1}{b} x=\frac{\ln(x)}{\ln\left(\frac{1}{b}\right)}=\frac{\ln(x)}{\ln(1)-\ln(b)}=\frac{\ln(x)}{-\ln(b)}$$

Answer 2

$$\log_{1/b} x=\frac{\log_b x}{\log_b \frac{1}{b}}=\frac{\log_b x}{-1}=-\log_b x$$

Answer 3

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:

$y= \log_{1/b}x = - \log_b(x).$