# how to simplify $\frac{\log_2 625}{\log_2 125}$

How can i simplify this:

$\dfrac{\log_2 625}{\log_2 125}$

Thanks

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There is nothing to solve, since there is no equation. Please put a little bit of effort in to make your question readable. –  Thomas Rot May 2 '11 at 7:09
@dramasea: if this is a homework, please tag it so. –  Thomas Connor May 2 '11 at 7:10
reformat already.. –  dramasea May 2 '11 at 7:11
Or even $\log(a^n)=n\log(a)$. –  Phira May 2 '11 at 7:19

Here it is:

$$\frac{\log_{2}625}{\log_{2}125} = \log_{125}625 = \frac{\log_{5}625}{\log_{5}125} = \frac{\log_{5}5^4}{\log_{5}5^3} = \frac{4}{3}$$

by virtue of a basic property of logarithms. The basic property I am referring to is the following:

$$\frac{\log_{a}x}{\log_{a}y} = \log_{y}x$$

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I'm asking $\frac{log_2 625}{log_2 125}$ –  dramasea May 2 '11 at 7:27
@dramasea: That's what he answered - he references the property that allows him to do that. –  mixedmath May 2 '11 at 7:30
@dramasea: Actually I use it twice: the first time, I use $\frac{log_{a}x}{log_{a}y} = log_{y}x$, and the second time I use the 'reverse': $log_{y}x = \frac{log_{a}x}{log_{a}y}$. Do you know this equality? –  Thomas Connor May 2 '11 at 7:33
It's called the Change of Base Property –  Nicolas Villanueva May 2 '11 at 7:34
Thanks for detail explaination, Thomas. –  dramasea May 2 '11 at 7:58

Without a change of base:

$$\frac{\log_2 625}{\log_2 125} =\frac{\log_2 \left( 5^4\right)}{\log_2 \left(5^3\right)} =\frac{4 \log_2 5}{3 \log_2 5} =\frac{4 }{3 } .$$

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@Henry: Consider to use hints. –  AD. May 2 '11 at 10:35
This is the simplest of all answers. Using a change of base formula and then writing everything as $5^4$ and so on, instead of just writing it as $5^4$ in the first place only adds unnecessary steps. –  Graphth May 2 '11 at 13:23
@AD.: There were already two answers (one accepted) giving $4/3$ when I replied –  Henry May 2 '11 at 15:25
@Numth: I couldn't agree more. –  Chris Leary May 2 '11 at 17:47
@Numth: I also agree, this is the most transparent answer - I gave the same comment on to the other writers too. –  AD. May 2 '11 at 19:41

Although you have now changed your question, the spirit of Thomas's answer is still correct. In particular, $\dfrac{log_2 625}{log_2 125} = \dfrac{log_5 625}{log_5 125} = \dfrac{4}{3}$. Does that make sense?

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@Thomas: Aha! I see you too have now changed your answer to his question. Perhaps I will beat you to it before he changes it next. –  mixedmath May 2 '11 at 7:23
;-) –  Thomas Connor May 2 '11 at 7:25
Consider to use hints. –  AD. May 2 '11 at 10:36