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

bnosnehpets
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1 Answers1

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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.

ajotatxe
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  • 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
  • I guess what I am asking is: why, when taking logs with bases less than 1, does the solution not work? – bnosnehpets May 20 '15 at 14:43
  • @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 00:26
  • @David thanks that makes it clearer – bnosnehpets May 21 '15 at 09:55