How does $\log_{1/b}x$ become $- \log_b(x).$?

Question

I understand the other case $$\log _x\left(\frac{1}{x}\right)=-1 $$ but when the log is to the base of a fraction, I don't know how to prove so it is true

A side remark: Be conscious that this kind of exercise is very academical: bases of logarithms that are $<1$ are never used in practice. — Jean Marie, Sep 12 '20 at 06:51
Does this answer your question? How to show that $- \log_b x = \log_{\frac{1}{b}} x$ — Arnaud D., Sep 14 '20 at 13:07

score 2 · Answer 1 · answered Sep 12 '20 at 06:49

2

Hint: $b^y=x\iff (1/b)^{-y}=x$.

answered Sep 12 '20 at 06:49

score 1 · Accepted Answer · answered Sep 12 '20 at 06:47

1

$$\log_x y=\frac{\log x}{\log y}$$ So $$\log_{1/b}x =\frac{\log x}{\log (1/b)}=\frac{\log x}{-\log b}=-\log_b x$$

answered Sep 12 '20 at 06:47

Z Ahmed

43,235

score 0 · Answer 3 · answered Sep 12 '20 at 06:56

0

Solving it for a very general case as follows:

Now to get your result, just put n=-1

answered Sep 12 '20 at 06:56

Soumyadwip Chanda

1,506

score 0 · Answer 4 · answered Sep 12 '20 at 07:16

0

By definition

$$y=\log_{1/b}x \iff\left(\frac 1 b\right)^y=x\iff \frac 1 {b^y}=x \iff\frac1x=b^y \iff x=b^{-y}\iff -y=\log_b x$$

answered Sep 12 '20 at 07:16

user

154,566

How does $\log_{1/b}x$ become $- \log_b(x).$?

4 Answers4