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So, I have a question here:

$\log_2 M + \log_3 N$

The goal is to simplify this as far as possible. I tried to do change of base, but that didn't really help. I thought of rearranging and manipulating it with the laws of log, namely the following:

$\log_a(bc)=\log_ab+\log_ac$

$\log_a(b/c)=\log_ab−\log_ac$

$\log_ab^n=n\log_ab$

$\log_ab=1/\log_ba$

$(\log_ab)(\log_bc)=\log_ac$

$\log_aa=1$

I tried a few methods, but I couldn't really make any progress. Can anyone help?

5201314
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user8594725
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  • Getting one single logarithmic term is not possible, but converting everything into log base 10, is. See answer below – imranfat May 20 '21 at 18:22

3 Answers3

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Use change of base formula on both given logarithms, and then make a common denominator: $\frac{\log M}{\log2}+\frac{\log N}{\log3}$=$\frac{1}{(\log2)(\log3)}$$(\log3\log M+\log2\log N)$

5201314
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imranfat
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You can try $$\log_2M + \log_3N=\frac{\log M}{\log2}+\frac{\log N}{\log3}=\log(M^{\frac{1}{\log2}})+\log(N^{\frac{1}{\log3}})=\log(M^{\frac{1}{\log2}}N^{\frac{1}{\log3}}).$$

zjs
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  • After the first $=$ sign, things go wrong. You used the change of base formula (the intend is ok) incorrectly – imranfat May 20 '21 at 18:17
  • What would you do to change that then? – user8594725 May 20 '21 at 18:18
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    @imranfat, actually the error is in the first equality; it should be $\log_2M+\log_3N=(\log_2e)\log M+(\log_3e)\log N$ (assuming $\log$ means $\log_e$). If you make that correction, then everything is OK. – Barry Cipra May 20 '21 at 22:43
  • @BarryCipra Yes you are right. My wording was poorly chosen, but when my eyes passed that first equal sign, I misswallowed and I stopped reading the rest. I think the intention was good though, the use of change of base. – imranfat May 20 '21 at 22:46
  • $\log_2M + \log_3N=\frac{\log M}{\log 2}+\frac{\log N}{\log 3}$ – Aaron Hendrickson May 20 '21 at 22:47
  • @imranfat, agreed, the basic idea is good. I had to check and doublecheck where the $\log_2e$ and $\log_3e$ belonged, since I usually get things wrong on the first attempt. – Barry Cipra May 20 '21 at 22:48
  • My bad! It should be corrected now. – zjs May 22 '21 at 05:00
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The basic idea is that $\log_a (b) =\dfrac{\log_c(b)}{\log_c(a)} $ holds for any $c$. So you can choose the most convenient $c$.

Also note that this implies that $\log_b(a)\log_a(b) = 1 $ so that $\log_a (b) =\log_c(b)\log_a(c) $.

So

$\begin{array}\\ \log_2 M + \log_3 N &=\dfrac{\log_c M}{\log_c 2} + \dfrac{\log_c N}{\log_c 3}\\ &=\log_c M\log_2 c + \log_c N\log_3 c\\ &=\log_c M^{\log_2 c} + \log_c N^{\log_3 c}\\ &=\log_c (M^{\log_2 c} N^{\log_3 c})\\ \end{array} $

Choose $c=2$ or $c=3$ to simplify this a little.

marty cohen
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