Find all real numbers satisfying the given equation.

algebra-precalculus

I came across a question which required solving two equations in real numbers $(x,y) $.
The two equations were:

\begin{align}
\log_3{x} +\log_2{y} &=2 \\
3^{x}-2^{y} &= 23
\end{align}

Now, an obvious solution is $(3,2) $. But I want to know that how do we actually solve such equations with both exponentials and logarithms simultaneously? I tried substituting the logarithmic terms but that gave me more complicated terms in the second equation which was hard to deal with. Please help.

Best Answer

If you increase $x$, the left-hand sides of both equations increase. If you increase $y$, the left-hand side of the first equation increases, while the left-hand side of the second equation decreases.

That means that if there is some other solution $(a,b)$ and, let's say $a>3$, then the first equation says that $b<2$ while the second equation says that $b>2$.

So there are no more solutions (at least not among the positive reals).

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