Simultaneous equations with logarithms? Given that $a$ and $b$ are positive constants and $a&gt;b$, solve the simultaneous equations.   

$a+b=13$
$\log_6 a +\log_6 b= 2$   

I have tried doing this but I can’t figure out what to do first. I am thinking that you take logs of equation $1$ but I’m not sure.
$\log_6a+\log_6b=\log_6(ab)$

$\log_6x=2$ if and only if $x=6^2=36$

Can you go on now, without peeking at the spoiler?


  \begin{cases}a+b=13\\ab=36\end{cases} has an obvious solution
Hint: Writing
$$\frac{\ln(a)}{\ln(6)}+\frac{\ln(b)}{\ln(6)}=2$$ with the first equation $$b=13-a$$ we get
$$\ln(a)+\ln(13-a)=2\ln(6)$$ Using the law of logarithm we get
$$\ln(a(13-a))=2\ln(6)$$
and $$a(13-a)=e^{2\ln(6)}$$
Can you finish?