$$\left( \frac{6}{7} \right) ^n < \frac{1}{65}$$ 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?
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.