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

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?

  • $\begingroup$ 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

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.

| cite | improve this answer | |
  • $\begingroup$ 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

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 $25^{\frac 32} = 125$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.