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Problem

$\dfrac{\log125}{\log25} = 1.5$

From my understanding, if two logs have the same base in a division, then the constants can simply be divided i.e $125/25 = 5$ to result in ${\log5} = 1.5$ but that is not the case as ${\log5} \neq 1.5$ .

Correct answer

Each log can be rewritten to be $\frac{3\log5}{2\log5} = 1.5$ therefore $\frac{3}{2} = 1.5$

I'm unsure why this is correct over the previous method.

My question

What was wrong with simply dividing the constants $125/25 = 5$ versus rearranging the logarithm?

Computing Corn
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2 Answers2

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Dividing logs which have the same base changes the base of the log.

That is $\frac {\log a}{\log b} = \log_b a$

It doesn't matter what base we were using on the left hand side. It will change the base of the log as above.

$\frac {\log 125}{\log 25} = \log_{25} 125$ and $25^{\frac 32} = 125$

Doug M
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Your "understanding" is just totally wrong. It's not true that $\frac{\log a}{\log b}=\log(a/b)$ in general, and indeed this problem is a counterexample.

Eric Wofsey
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  • The confusion stems from the fact that $log6 - log3 = \dfrac{\log6}{\log3} = log2$ . I believe you can divide $ \dfrac{\log6}{\log3}$ to result in $ log2$ hence I would have thought you can apply $\dfrac{\loga}{\logb} = log(a/b)$ – Computing Corn Jun 26 '18 at 22:51
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    No, that's false as well. $\log 6-\log 3=\log(6/3)=\log 2$ but this is not the same as $\frac{\log 6}{\log 3}$. – Eric Wofsey Jun 26 '18 at 22:52
  • Ah that makes sense. I was getting confused with the minus operation and division. Thanks for clearing it up! – Computing Corn Jun 26 '18 at 22:53