Solve the following system of equations, $\log_{9}x = \log_{12}y = \log_{15}(x + y)$, and more generally, $\log_ax = \log_by = \log_c(x \pm y)$. 
Solve the following system of equations $$\log_{9}x = \log_{12}y = \log_{15}(x + y)$$

This isn't really just a question about the above system, but more of a question asking for the general method for solving $\log_ax = \log_by = \log_c(x \pm y)$ where $a, b, c \in \mathbb (0, +\infty) \setminus \{1\}$, as I've seen this specific type of system more times than to be comfortable not knowing how to tackle it. In this case, it just so happens to be that $9^2 + 12^2 = 15^2$, which means the roots of this system are $x = 9^2 = 81$ and $y = 12^2 = 144$.
But let's pretend for a moment that I didn't know that was the case. How should I go about solving this normally, especially by hand and under a time constraint of 5 minutes? This isn't even considered entirely solvable by WolframAlpha, or at least, that's my interpretation of its output when given the above system.
Anyhow, we have that $\dfrac{\ln x}{\ln 9} = \dfrac{\ln y}{\ln 12} = \dfrac{\ln (x + y)}{\ln 15}$, from which it can be obtained that $$\left\{ \begin{aligned} x &= y^{\log_{12}{15}} - y\\ y &= x^{\log_{9}{15}} - x \end{aligned} \right. \implies \left\{ \begin{aligned} \dfrac{\ln x}{\ln 9} &= \dfrac{\ln(x^{\log_{9}{15}} - x)}{\ln 12}\\ \dfrac{\ln y}{\ln 12} &= \dfrac{\ln(y^{\log_{12}{15}} - y)}{\ln 9} \end{aligned} \right. \iff \left\{ \begin{aligned} \dfrac{\ln x}{\ln 15} &= \dfrac{\ln(x^{\log_{9}{5/3}} - 1)}{\ln 12}\\ -\dfrac{\ln y}{\ln 15} &= \dfrac{\ln(y^{\log_{12}{5/4}} - 1)}{\ln 9} \end{aligned} \right.$$
I can't lie, I don't know what to do next. It'd be wonderful if you could help me out, thanks for reading~
 A: Let $t=\log_9x=\log_{12}y=\log_{15}(x+y)$. That means that $x=9^t, y=12^t, x+y=15^t$, i.e.
$$9^t+12^t=15^t$$
Or (divide both sides by $15^t$):
$$\left(\frac{3}{5}\right)^t+\left(\frac{4}{5}\right)^t=1$$
Now, the left side is a sum of two strictly decreasing functions, so this equation has at most one solution in $t$, and it is easily checked that $t=2$ is a solution (remember a Pythagorean $3:4:5$ triangle). From there, $x=81, y=144, x+y=225$.
It is to be appreciated that the insight “$t=2$ is ‘obviously’ a solution” is a fundamental part of finding the exact solution under time pressure. Otherwise, $(3/4)^t+(4/5)^t=1$ is a transcendental equation, of the type which can in most cases be solved only numerically.
A: Writing the system of equations $$\log_a(x) = \log_b(y) = \log_c(x + y)=k$$, we have $x=a^k$, $y=b^k$ and we need to find the zero of function
$$f(k)=\log \left(a^k+b^k\right)-k \log (c)$$ which does not show explicit solutions. So, in the most general case, a numerical method would be required.
There is one thing very useful is that the function is bounded by $\big[\log \left(2a^k\right)-k \log (c)\big]$ and  $\big[\log \left(2b^k\right)-k \log (c)\big]$ that is to say that the solution $k$ is
$$\min \left(\frac{\log (2)}{\log (c)-\log (a)},\frac{\log (2)}{\log (c)-\log
   (b)}\right) < k < \max \left(\frac{\log (2)}{\log (c)-\log (a)},\frac{\log (2)}{\log (c)-\log
   (b)}\right)$$ For your example $(a=9,b=12,c=15)$, it gives $1.35692<k<3.10628$.
Between the bounds, $f(k)$ will be close to linearity and this is very good for any solver. Using
$$k_0=\frac{1}{2} \left(\frac{\log (2)}{\log (c)-\log (a)}+\frac{\log (2)}{\log (c)-\log(b)}\right)$$ Newton iterates would be
$$\left(
\begin{array}{cc}
n & k_n \\
 0 & 2.2315996 \\
 1 & 1.9984339 \\
 2 & 1.9999999 \\
 3 & 2.0000000
\end{array}
\right)$$
Using a less trivial case $(a=7,b=13,c=19)$
$$\left(
\begin{array}{cc}
n & k_n \\
 0 & 1.2603466 \\
 1 & 1.0843855 \\
 2 & 1.0865853 \\
 3 & 1.0865856
\end{array}
\right)$$
Edit
Suppose that $b>a$ and rewrite
$$f(k)=\log(b^k)+\log \left(1+\left(\frac{a}{b}\right)^k\right)-k \log(c)$$ and we want to solve for $k$ the equation
$$\log \left(1+\left(\frac{a}{b}\right)^k\right)=k \log \left(\frac{c}{b}\right)$$ Approximating, solve
$$\left(\frac{a}{b}\right)^k=k \log \left(\frac{c}{b}\right)\implies k_0=\frac{W\left(\frac{\log \left(\frac{b}{a}\right)}{\log
   \left(\frac{c}{b}\right)}\right)}{\log \left(\frac{b}{a}\right)}$$ where appears Lambert function.
Make one iteration of Newton method
$$k_1=k_0-\frac{\log \left(a^{k_0}+b^{k_0}\right)-{k_0} \log (c)}{\frac{a^{k_0} \log (a)+b^{k_0} \log
   (b)}{a^{k_0}+b^{k_0}}-\log (c)}$$
For $(9,12,15)$, $k_1=1.99724$ and for $(7,13,19)$, $k_1=1.08508$
