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

today I'm in doubt on calculating the follow expression $\log_4 3 * \log_9 32$

Changing all to base 4: Working on: $\log_4 3 * \dfrac{\log_4 32}{\log_4 9}$

Ending with: $\log_4 3 * \dfrac{2 + \log_4 2}{2*\log_4 3}$

There's a way to simplify it more ? Also, do you know any resource explaining more on rules on every kind of operation with logs ?

Thanks in advance

share|cite|improve this question
up vote 6 down vote accepted

$\log_4 2$ is $\frac{1}{2}$. So it evaluates to $\frac{5}{4}$.

share|cite|improve this answer
Thanks, I've evaluated it to $\log_4 3 * \dfrac{\frac{5}{4}}{2*\log_4 3}$ – aajjbb Nov 15 '12 at 17:43
No. The whole expression evaluates to... wait did you mean something else by chance? I mean could you put brackets on your actual problem so that we can read it correctly. I am wondering why u didnt cancel $\log_4 3$. Also the numerator is $\frac{5}{2}$. I thought you could cancel the $\log_4 3$ and you would get $\frac{5}{4}$. But apparently your question was something different... – Gautam Shenoy Nov 15 '12 at 17:48
Sorry, you're right, I have forgotten to cancel the $\log_4 3$ from the expression, so the final result will be $\dfrac{5}{4}$ – aajjbb Nov 15 '12 at 18:11

this evaluation is based on following basic rules

$\log_c ab=\log_ca+\log_cb$

$\log_c a^n=n\log_ca$


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.