Consider the following equation: $3^x=27$ This is a very simple exponential problem. To solve this, we just take the log of both sides (we could've just used the common base method, but for the sake of this question, let's use the logarithm method). Therefore, $\log3^x=\log27$ $x*\log3=\log27$ $x=\frac{\log27}{\log3}$ At this point, the base are the same (10). The division rule for logarithm states that if the bases are the same, we can subtract the arguments. But if I do that, I get $\log24$ which is not the same as the answer $3$. But when I typed $\frac{\log27}{\log3}$, it gets me 3. So, why this rule fails to work?

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    $\begingroup$ No. The property of logarithms is that $\log(a) - \log(b) = \log(\frac{a}{b})$: the difference of logarithms is the logarithm of the quotient. You are trying to do $\log(a-b)=\frac{\log(a)}{\log(b)}$, and that is just not true at all. $\endgroup$ Feb 9 at 3:06
  • $\begingroup$ $\log 27 = \log 3^3 = 3\log 3$. $\endgroup$ Feb 9 at 3:12

1 Answer 1


I think you are confusing $$\log a -\log b = \log(\frac{a}{b})$$which is the actual rule, for $$\frac{\log(a)}{\log(b)} = \log(a-b) $$ which is wrong. Here one would use the change of base identity which is $$\frac{\log a}{\log b} = \log_ba $$


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