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.

  • $\begingroup$ 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 $\endgroup$ – Don Thousand Mar 27 at 15:23
  • $\begingroup$ I'm sorry but I still don't get it. Would you please give a little more hint? $\endgroup$ – 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 $b>2$.

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

  • $\begingroup$ I upvoted, but you can restrict to positive reals since $\log_a x$ is undefined in the reals for nonpositive reals. $\endgroup$ – InterstellarProbe Mar 27 at 15:28
  • $\begingroup$ Thank you very much. The solution was great. $\endgroup$ – Shashwat1337 Mar 27 at 15:28
  • $\begingroup$ @InterstellarProbe I thought I made that restriction clear enough. Do you want me to make it clearer? $\endgroup$ – Arthur Mar 27 at 15:29
  • $\begingroup$ @Arthur, I read the last statement as saying there might be solutions among the nonpositive reals. $\endgroup$ – InterstellarProbe Mar 27 at 15:42
  • 1
    $\begingroup$ @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. $\endgroup$ – InterstellarProbe Mar 27 at 16:08

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