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

How can i simplify this:

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

Thanks

-
reformat already.. – dramasea May 2 '11 at 7:11
You can use $log(a\cdot b)=log(a)+log(b)$ to solve this. – Rasmus May 2 '11 at 7:14
can do it?? i dont cant get it – dramasea May 2 '11 at 7:18
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:

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

-
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:

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

-
@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?

-
@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