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The question says if $\log_6(2)$ is $a$ and $\log_5(3)$ is $b$, express $\log_5(2)$ in terms of $a$ and $b$.

I have tried the change of base formula for $ab$ to no avail, can someone give me a hint to get started, and the solution, hidden behind a spoiler tag, if I get stuck? Thanks.

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Write down the expressions for $a$ and $b$ and define your target as $c$: $$a=\log_6(2)=\frac{\ln(2)}{\ln(6)}$$ $$b=\log_5(3)=\frac{\ln(3)}{\ln(5)}$$ $$c=\log_5(2)=\frac{\ln(2)}{\ln(5)}$$ Now use $\ln(6) = \ln(2) + \ln(3)$ and get $$c=\frac{\ln(2)}{\ln(5)} = \frac{\ln(2)}{\ln(6)} \frac{\ln(6)}{\ln(5)} =\frac{\ln(2)}{\ln(6)} \frac{\ln(2)+\ln(3)}{\ln(5)}$$ $$c= \frac{\ln(2)}{\ln(6)} \frac{\ln(2)}{\ln(5)} + \frac{\ln(2)}{\ln(6)} \frac{\ln(3)}{\ln(5)}=ac + a b$$ Now solve for $c$ and get the solution $$c=\frac{ab}{1-a}\cdot$$ And to be sure you can do a sanity check $$a=\log_6(2)\approx 0.386853$$ $$b=\log_5(3)\approx 0.682606$$ $$c=\log_5(2)\approx 0.430677$$ $$\frac{ab}{1-a}\approx\frac{0.386853\times0.682606}{1-0.386853}\approx 0.430676785\dots \approx c$$

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