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How can i simplify this:

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

Thanks

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    Please reformat your question.2011-05-02
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    reformat already..2011-05-02
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    You can use $log(a\cdot b)=log(a)+log(b)$ to solve this.2011-05-02
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    can do it?? i dont cant get it2011-05-02
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    Or even $\log(a^n)=n\log(a)$.2011-05-02
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    @dramasea: please don't do what you did on this question again. The latter edits you made completely changed the meaning/character of the question, which would've invalidated other users' helpful answers.2011-05-02

3 Answers 3

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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.2011-05-02
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    ;-)2011-05-02
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    Consider to use hints.2011-05-02
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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.2011-05-02
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    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.2011-05-02
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    @AD.: There were already two answers (one accepted) giving $4/3$ when I replied2011-05-02
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    @Numth: I couldn't agree more.2011-05-02
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    @Numth: I also agree, this is the most transparent answer - I gave the same comment on to the other writers too.2011-05-02
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    @Henry: Your answer is the best one - however I am sure you can write that answer in hints..2011-05-02
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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}$2011-05-02
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    @dramasea: That's what he answered - he references the property that allows him to do that.2011-05-02
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    @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?2011-05-02
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    It's called the [Change of Base Property](http://en.wikipedia.org/wiki/Logarithmic_identities#Changing_the_base)2011-05-02
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    Thanks for detail explaination, Thomas.2011-05-02
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    @Thomas Connor: Consider to use hints.2011-05-02