Simultaneous equations with logarithms? Given that $a$ and $b$ are positive constants and $a>b$, 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. 
 A: $\log_6a+\log_6b=\log_6(ab)$
$\log_6x=2$ 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

A: It comes down to finding two numbers, given their sum $s$ and their product $p$ :
$$\begin{cases}a+b=13\\ \log_6 a+\log_6 b=2\end{cases}\iff\begin{cases}a+b=13\\ \log_6ab=2\end{cases}\iff\begin{cases}a+b=13\\ ab=6^2=36\end{cases}$$
so $a$ and $b$, by the theory of quadratic equations, are the roots of
$$x^2-sx+p=x^2-13x+36=0.$$
You can find the roots applying the rational roots theorem.
A: Taking the logarithm of a sum leads you nowhere. But taking the antilogarithm yields a product.
$$6^{\log_6a+\log_6b}=6^{\log_6a}6^{\log_6b}=ab=6^2=36.$$
You have reduced to a problem where the sum and product of two numbers is known. By the Vieta formulas, these are the solutions of the quadratic equation
$$x^2-13x+36=0.$$
A: 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?
