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Sorry for such a simple question.

I have $y\ln(5) = 2\ln 3$

And I wanted to know if I can solve for $y$ by just dividing the logarithm on both sides?

So I would get $y = \frac {2\ln 3}{\ln(5)} = 2 \ln 3 - \ln 5$

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    $\begingroup$ $y = \frac {2\ln 3}{\ln(5)} \color{red} {\neq 2 \ln 3 - \ln 5}$ $\endgroup$ – Claude Leibovici May 11 at 6:04
  • $\begingroup$ why not? i thought that was the property of logarithms $\endgroup$ – user130306 May 11 at 6:05
  • $\begingroup$ No, it is not. Check again. $\endgroup$ – A. Pongrácz May 11 at 6:05
  • $\begingroup$ Logarithm of a quotient is the difference of logarithms. That is not the same as “quotient of logarithms is the difference of logarithms”. $\endgroup$ – Arturo Magidin May 11 at 6:09
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To more directly answer your question, yes, you can divide both sides by $\ln(5)$ to isolate $y$.

However also as pointed out in other posts, you're mixing up properties. $\ln(a/b) = \ln(a)-\ln(b)$ - it's not $\ln(a)/\ln(b)=\ln(a)-\ln(b)$. For an easy way to see this, take $a=b=1$. Then

$$\ln\left( \frac 1 1 \right) = \ln(1) = 0 = 0 - 0 = \ln(1) - \ln(1)$$

but

$$\frac{\ln(1)}{\ln(1)} = \frac 0 0 \ne \ln(1) - \ln(1) = 0$$

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You can only use that $$\ln\left(\frac{a}{b}\right)=\ln(a)-\ln(b)$$ for $$a,b>0$$

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Note clearly the difference between $\frac {log (a)} {log(b)} $and $log (\frac a b )$: $\frac {log (a)} {log(b)} ≠log (\frac a b )= log (a)-log(b)$

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