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I'm working on a recurrence problem which needs me to simplify $\log_{60/47}3$?

Since $\frac{60}{47}\approx 1.2765$, how would I know how much of that I need to multiply by itself to get $3$? I have no problem simplifying log of 8 base 3 $\log_28$, $\log_39$ or something similar in difficulty, but I'm not sure how to solve this one.

Help!

Gannicus
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    How have you calculated $\log_38=?$ – lab bhattacharjee Jul 25 '13 at 03:20
  • @Gannicus You meant $\log_39$ or $\log_28$? I'll edit that for you since it makes much more sense ... anyway –  Jul 25 '13 at 03:21
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    If you can calculate $\log_3 8$, would the same methods help you calculate $\log_360$ and $\log_347$? If so, perhaps you would like to try using $\log_{60/47}3=\dfrac{1}{\log_3(60/47)}=\dfrac{1}{\log_360-\log_3 47}$. – Jonas Meyer Jul 25 '13 at 03:22
  • @Danny: You may be right, but can we let Gannicus answer before changing the content? I am not so sure. – Jonas Meyer Jul 25 '13 at 03:29

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The smart play here is to write this as an algebraic equation. If $x = \log_\frac{60}{47}3$, then we have $$ \left(\frac{60}{47}\right)^x=3 $$ Now, choose your favorite logarithm (i.e. choose any base) and apply it to both sides. Then, use the log rules: $$ \log\left[\left(\frac{60}{47}\right)^x\right]=\log3\\ x\log\left(\frac{60}{47}\right)=\log3\\ x = \frac{\log 3}{\log\left(\frac{60}{47}\right)}=\frac{\log 3}{\log 60-\log47} $$ Then, just plug it into a calculator and see what you get. For example, google gives you the answer $4.4988$ using logs base $10$.

Ben Grossmann
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    If the plan is to plug into a calculator, one needs only the identity $\log_b a=\dfrac{\log a}{\log b}$ in the first place, although I suppose it does no harm to show a way to derive the identity. But $\log(60/47)$ is easier to enter into my calculator directly than $\log 60-\log 47$. – Jonas Meyer Jul 25 '13 at 03:32
  • @JonasMeyer that's a fair point, I could have left it like that. Everything having been said though, I'm not really sure what OP is going for, but I gave it a shot. – Ben Grossmann Jul 25 '13 at 03:55
  • It does not matter what base you use, since the answer is the ratio of logs to the same base. – marty cohen Jul 25 '13 at 04:35
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    @martycohen hence the "pick your favorite logarithm" – Ben Grossmann Jul 25 '13 at 05:38