# Conversion of bases with logarithms The question says if log_6(2) is a and log_5(3) is b, express

Conversion of bases with logarithms
The question says if ${\mathrm{log}}_{6}\left(2\right)$ is $a$ and ${\mathrm{log}}_{5}\left(3\right)$ is $b$, express ${\mathrm{log}}_{5}\left(2\right)$ 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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autosmut6p
Write down the expressions for $a$ and $b$ and define your target as $c$:
$a={\mathrm{log}}_{6}\left(2\right)=\frac{\mathrm{ln}\left(2\right)}{\mathrm{ln}\left(6\right)}$
$b={\mathrm{log}}_{5}\left(3\right)=\frac{\mathrm{ln}\left(3\right)}{\mathrm{ln}\left(5\right)}$
$c={\mathrm{log}}_{5}\left(2\right)=\frac{\mathrm{ln}\left(2\right)}{\mathrm{ln}\left(5\right)}$
Now use $\mathrm{ln}\left(6\right)=\mathrm{ln}\left(2\right)+\mathrm{ln}\left(3\right)$ and get
$c=\frac{\mathrm{ln}\left(2\right)}{\mathrm{ln}\left(5\right)}=\frac{\mathrm{ln}\left(2\right)}{\mathrm{ln}\left(6\right)}\frac{\mathrm{ln}\left(6\right)}{\mathrm{ln}\left(5\right)}=\frac{\mathrm{ln}\left(2\right)}{\mathrm{ln}\left(6\right)}\frac{\mathrm{ln}\left(2\right)+\mathrm{ln}\left(3\right)}{\mathrm{ln}\left(5\right)}$
$c=\frac{\mathrm{ln}\left(2\right)}{\mathrm{ln}\left(6\right)}\frac{\mathrm{ln}\left(2\right)}{\mathrm{ln}\left(5\right)}+\frac{\mathrm{ln}\left(2\right)}{\mathrm{ln}\left(6\right)}\frac{\mathrm{ln}\left(3\right)}{\mathrm{ln}\left(5\right)}=ac+ab$
Now solve for $c$ and get the solution
$c=\frac{ab}{1-a}\cdot$
And to be sure you can do a sanity check
$a={\mathrm{log}}_{6}\left(2\right)\approx 0.386853$
$b={\mathrm{log}}_{5}\left(3\right)\approx 0.682606$
$c={\mathrm{log}}_{5}\left(2\right)\approx 0.430677$
$\frac{ab}{1-a}\approx \frac{0.386853×0.682606}{1-0.386853}\approx 0.430676785\cdots \approx c$
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