Solving for the values of (x+y) upto 3 decimal places.

Question

Given a equation:
$72^x48^y=6^{xy}$ where $x,y\in \Bbb Q \ne 0$
Find the value of $(x+y)$

I tried using $\log$ but always end up with 2 variables $x,y$.
Thanks for your help

Answer 1

Apply logarithms, so $$ x\ln 72+y\ln 48=xy\ln 6$$ or by expanding the integers $72=2^33^2$, $49=2^43^1$, $6=2^13^1$ $$ 3x\ln 2+2x\ln3+4y\ln 2+y\ln 3=xy\ln 2+xy\ln3$$ As $\ln2$ and $\ln3$ are rationally independent and $x,y$ are rational, this gives us in fact two equations $$\begin{align}3x+4y&=xy\\2x+\hphantom{0}y&=xy\end{align} $$ By equating the expressions for $xy$, $3x+4y=2x+y$, i.e., $$ x=-3y.$$ Substituting this gives us $-5y=-3y^2$, hence either $y=0$ (and then $x=0$, but that solution was excluded) or $$y=\frac 53\qquad\text{and}\qquad x=-5.$$