Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

I'm in doubt on simplifying the expression: $\log_2 6 - \log_4 9$

Working on it I've got: $\log_2 6 - \dfrac{\log_2 9}{2}$

There's anyway to simplify it more ? I'm learning logarithms now so I'm not aware of all properties and tricks.

Thanks in advance

share|cite|improve this question
If you put a forward slash in front of log, you get $\log$ rather than $log$. – Daniel Littlewood Nov 18 '12 at 20:47
Thanks, corrected; – aajjbb Nov 18 '12 at 20:58
up vote 2 down vote accepted

You can do a bit more simplification. The important properties of the log function are, for any base $a>0$,

  • $\log_a(bc)=\log_ab+\log_ac$, so, for example, $\log_26=\log_22+\log_23$
  • $\log_a(b/c)=\log_ab-\log_ac$
  • $\log_ab^n=n\log_ab$
  • $\log_ab=1/\log_ba$
  • $(\log_ab)(\log_bc)=\log_ac$
  • $\log_aa=1$

So, for example, we can simplify $\log_26-\log_49$ as $$ \begin{align} \log_26-\log_49&=\log_2(2\cdot3)-\log_4(3^2) \\ &= \log_22+\log_23-2\log_43 &\text{using the first and third identities}\\ &=1+\log_23-2\log_43 &\text{using the sixth identity}\\ &=1+\log_23-2\log_42\log_23 &\text{using the fifth}\\ &=1+\log_23-2(\log_23)/\log_24 &\text{using the fourth}\\ &=1+\log_23-2(\log_23)/2 &\text{using the third}\\ &=1+\log_23-\log_23 &\text{using a bit of algebra}\\ &=1 \end{align} $$

share|cite|improve this answer
Thank you, this table will helps me a lot. but in the Latex representation, the base is being shadowed by the 'G' of log. – aajjbb Nov 18 '12 at 21:10
@aajbb Yeah, that happens to a lot of people, myself included. The consensus is that this site's LaTeX processor doesn't play nicely with some browsers. Often, if you simply tell the browser to refresh the page, the problem with overlapping text goes away. – Rick Decker Nov 18 '12 at 21:16

You're correct so far. Bring out the factor of $\frac{1}{2}$ to get $\frac{1}{2}(2\log_{2}(6)-\log_{2}(9))$. Since $a\log(b)=\log(b^{a})$ $\log(a)-\log(b)=\log(\frac{a}{b})$, you get $2\log_{2}(6)=\log_{2}(6^{2})$, and $$\frac{1}{2}(\log_{2}(36)-\log_{2}(9))=\frac{1}{2}\left(\log_{2}\left(\frac{36}{9}\right)\right)=\log_{2}(4)/2=2/2=1 $$

share|cite|improve this answer


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.