# Find all real numbers satisfying the given equation.

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.

• Raise the first equation to the third power, and take $\log_3$ of the second. You should be able to solve for $y$ with that – Don Thousand Mar 27 at 15:23
• I'm sorry but I still don't get it. Would you please give a little more hint? – Shashwat1337 Mar 27 at 15:26

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 $$b>2$$.
• I upvoted, but you can restrict to positive reals since $\log_a x$ is undefined in the reals for nonpositive reals. – InterstellarProbe Mar 27 at 15:28
• @Arthur, I understand, but suppose it could be well-defined. For all $x\le 0$, we have $\log_3 x \notin \mathbb{R}$. So, if $x\le 0, y>0$ or $x>0, y\le 0$, you have $\log_3 x + \log_2 y \notin \mathbb{R}$. Therefore, if a solution were to exist over nonpositive real numbers, you need $x<0,y<0$ (since the logarithm cannot be defined at zero without including a point at infinity where sums are not really possible). Now, for all $x<0, y<0$, you have $0<3^x<1$ and $0<2^y<1$ which means $-1<3^x-2^y<1$, which does not put it anywhere close to 23. So, negative solutions are not possible. – InterstellarProbe Mar 27 at 16:08