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How does one solve a logarithmic expression where the base is a fraction? In my example I am trying to solve the following:

$$ n^{\log_\frac{3}{2}(1)} \tag{1} $$

This is related to using the "master theorem" to solve recurrence relations. People usually give examples where they solve something like:

$$ {\log_\frac{1}{3}(27)} \tag{2} $$

which seems easy to understand. However, no one really gives an example of how one would go about solving the expression listed above (1).

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    $\begingroup$ In any base, $\log1$ is always $0$. $\endgroup$ – Allawonder Jan 9 '18 at 14:58
  • $\begingroup$ in any case you can change you base using formula $\log_a b=\frac{\ln b}{\ln a}=\frac{\log_c b}{\log_c a}$ $\endgroup$ – aid78 Jan 9 '18 at 14:59
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First evaluate the exponent. Here, the log of $1$ to any base is $0$, so you have $n^0=1$. If you had $n^{\log_{\frac 32}\frac 94}$ the exponent would be ${\log_{\frac 32}\frac 94}=2$ so you would have $n^2$

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  • $\begingroup$ Can you elaborate in terms of the numbers you came about? Like give a more specific example where 1 would not be included? Something like exchanging the 1 with 27. $\endgroup$ – atiesh Jan 9 '18 at 15:04
  • $\begingroup$ I did give an example in the last sentence. For $\log_{\frac 32}(27)$ you use the laws of logarithms $\log_{\frac 32}(27)=\frac {\log(27)}{\log(\frac 32)}$. I thought you knew that from your last example. $\endgroup$ – Ross Millikan Jan 9 '18 at 15:07
  • $\begingroup$ Oh! I see. Thank you very much for your help :D $\endgroup$ – atiesh Jan 9 '18 at 15:10
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You can always change bases with the formula

$$\log_a x = \frac{\log_b x}{\log_b a}.$$

So change to your favorite integer base:

$$\log_{3/2}1 = \frac{\log_2 1}{\log_2 3/2} =\frac{\log_2 1}{\log_2 3 -1}.$$

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