# Is this manipulation with logs allowed?

$$\left( \frac{6}{7} \right) ^n < \frac{1}{65}$$ The answer is, by looking at which way the sign should be round:

$$n > \log_\frac{6}{7}{\left(\frac{1}{65}\right)} \implies n>\frac {\log{\frac{1}{65}}}{\log{\frac{6}{7}}}$$

However if I try to solve it by taking logs of both sides: $$n\log{\frac{6}{7}} < \log{\frac{1}{65}}$$ When I divide by $\log{\frac{6}{7}}$ however, the sign switches only if it is less than $0$. Depending on the base of the log, the answer will either be (for example if the base is $\sqrt{\frac{6}{7}}$)... $$n<\frac {\log{\frac{1}{65}}}{\log{\frac{6}{7}}}$$ Or... $$n>\frac {\log{\frac{1}{65}}}{\log{\frac{6}{7}}}$$ ...which is correct. So what's wrong with my manipulation? Is there a formal way to show that the answer is indeed the second, by using the second method?

Edit:

I guess the main point of this question is to ask what is allowed when taking logs of both sides of an equation. I assumed these logs could be any base, but some bases seem to yield incorrect answers (for example what I wrote in the comments). Is there any way to formally show why this does not work?

• You should keep in mind that $\log x<0$ if $x<1$. In you case, $\log\frac1{65}<0$ and also $\log\frac67<0$. – Tom-Tom May 20 '15 at 14:22
• What's wrong with: $\log_{0.5}{0.25}=2$? @Tom-Tom – bnosnehpets May 20 '15 at 14:30
• @Tom-Tom I mean, assuming the log can be any base (can't it?)... – bnosnehpets May 20 '15 at 14:45

If you take $\log_a$, there are two possibilities.
If $a>1$ then $\log_a$ is increasing, and the direction of the inequality does not change.
If $0<a<1$, the function $\log_a$ is decreasing, and the direction of the inequality is reversed.
• How about in the example: $$n\log_{0.5}{\frac{1}{4}}<\log_{0.5}{\frac{1}{16}}$$ Then $2n<4$ and $n<2$. No? @ajotatxe – bnosnehpets May 20 '15 at 14:27
• @bnosnehpets If $x<y$, then $\log_{0.5} x > \log_{0.5} y$. The step where you take logs of both sides and get $n\log_a{\frac{6}{7}} < \log_a{\frac{1}{65}}$ is correct only if $a > 1.$ That step is a mistake if $a=0.5$. – David K May 21 '15 at 0:26