Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is an awkward question to me, it was not covered in class.


$$\begin{align*} \log_b 2 &= A, \\ \log_b 3 &= B, \\ \log_b 5 &= C. \end{align*} $$

Then, use the change-of-base formula to evaluate

$$\log_{b^2} 5$$


$$\log_{\sqrt b} 2 .$$

As an example answer, we're given $\log_3 b$ becomes $1/B$.

Hopefully someone understands where they're coming from.

share|cite|improve this question
$\log_{b^2}5=\dfrac{\log_b 5}{\log_b b^2}$; and $\log_{\sqrt b}2=\dfrac{\log_b 2}{\log_b \sqrt b}$... in general, $\log_c a=\dfrac{\log_b a}{\log_b c}$. – J. M. Nov 29 '11 at 2:24
up vote 3 down vote accepted

$\log_x y = \frac{\ln y}{\ln x}$


$\log_{b^2} 5 = \frac{\ln 5}{\ln b^2} = \frac{\ln 5}{2 \ln b}$

and you should be able to continue from here.

Edit: I just realized that you may not know what $\ln$ is. In which case think of it as $\log_e$ where $e$ is a well known number $e=2.71828...$.

share|cite|improve this answer


$$\log_{\sqrt b}2=\frac{\log_b2}{\log_b\sqrt b}=\frac{\log_b2}{\frac12\log_bb}=2A$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.