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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?

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  • $\begingroup$ In which basis is $\log 5$ equal to $1.5$? $\endgroup$ – Bernard Jun 26 '18 at 22:41
  • $\begingroup$ Do you understand why it's correct to say that $\log 125=3\log 5$? $\endgroup$ – Eric Wofsey Jun 26 '18 at 22:42
  • $\begingroup$ @Bernard It isn't . Sorry if it was not clear in the question as I was unable to perform the 'not equals to' symbol $\endgroup$ – Computing Corn Jun 26 '18 at 22:43
  • $\begingroup$ The code is \neq, as in LaTeX. $\endgroup$ – Bernard Jun 26 '18 at 22:43
  • $\begingroup$ @EricWofsey Yes I understand that you can remove the multiple by converting it to an exponent and vice versa. $\endgroup$ – Computing Corn Jun 26 '18 at 22:44
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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.

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  • $\begingroup$ 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 $\dfrac{\loga}{\logb} = log(a/b)$ $\endgroup$ – Computing Corn Jun 26 '18 at 22:51
  • $\begingroup$ No, that's false as well. $\log 6-\log 3=\log(6/3)=\log 2$ but this is not the same as $\frac{\log 6}{\log 3}$. $\endgroup$ – Eric Wofsey Jun 26 '18 at 22:52
  • $\begingroup$ Ah that makes sense. I was getting confused with the minus operation and division. Thanks for clearing it up! $\endgroup$ – Computing Corn Jun 26 '18 at 22:53
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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 $25^{\frac 32} = 125$

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