How to find x and y? Doesn’t there need to be 2 equations?

Using Newton-Raphson method, it gives the following solution:

x = 1/2   and     y=1/2

$2^{x+y}\,\cdot\,3^{x-y}\,\cdot\,6^{2x+2y}\ =\ 72\\~\\ 2^{x+y}\,\cdot\,3^{x-y}\,\cdot\,6^{2(x+y)}\ =\ 72\\~\\ 2^{x+y}\,\cdot\,3^{x-y}\,\cdot\,(6^2)^{x+y}\ =\ 72\\~\\ 2^{x+y}\,\cdot\,3^{x-y}\,\cdot\,36^{x+y}\ =\ 72\\~\\ 2^{x+y}\,\cdot\,36^{x+y}\,\cdot\,3^{x-y}\ =\ 72\\~\\ (2\,\cdot\,36)^{x+y}\,\cdot\,3^{x-y}\ =\ 72\\~\\ 72^{x+y}\,\cdot\,3^{x-y}\ =\ 72$

At this point we can see that the equation will definitely be satisfied if

x + y  =  1    and

x – y  =  0

So    x  =  1/2    and    y = 1/2    is definitely a solution.

But that doesn't necessarily mean it is the only solution.

I don't know how to prove that it is the only rational solution, but it does seem to be.

Maybe someone else can explain why that is.

LHS:

$\displaystyle 2^{x+y}.3^{x-y}.6^{2x+2y}=2^{x+y}.3^{x-y}.(2\times3)^{2x+2y}=2^{x+y}.3^{x-y}.2^{2x+2y}.3^{2x+2y}\\=2^{3x+3y}.3^{3x+y}$

RHS:

$\displaystyle 72=2^{3}.3^{2}$

It's an identity, so equating indices,

$\displaystyle 3x+3y=3,\\3x+y=2.$

Solve, x = 1/2, y = 1/2.

The solution is unique.

The expression p₁^Q₁*p₂^Q₂*....*p_n^Q_n for p₁,....,p_n distinct primes and Q₁,....,Q_n rational numbers will result in an integer if and only if Q₁,...,Q_n are all nonnegative integers