# How can I isolate a variable in a logarithmic equation?

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 $$y = \frac {2\ln 3}{\ln(5)} = 2 \ln 3 - \ln 5$$

• $y = \frac {2\ln 3}{\ln(5)} \color{red} {\neq 2 \ln 3 - \ln 5}$ – Claude Leibovici May 11 at 6:04
• why not? i thought that was the property of logarithms – user130306 May 11 at 6:05
• No, it is not. Check again. – A. Pongrácz May 11 at 6:05
• Logarithm of a quotient is the difference of logarithms. That is not the same as “quotient of logarithms is the difference of logarithms”. – Arturo Magidin May 11 at 6:09

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

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

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

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