$$2^{ x+y }=16\\ 3^{ x }-3^{ y }=24$$

Steps I took:

$$\ln(2^{ x+y })=\ln(2^{ 4 })$$

$$(x+y)\ln(2)=4\ln(2)\Rightarrow x+y=4$$

$$\ln(3^{ x })-\ln(3^{ y })=\ln(24)\Rightarrow x-y=\frac { \ln(3)+\ln(8) }{ \ln(3) } $$

I tried to simplify these equations in a way that would make them easier to work with. I got stuck at the last step. I don't even if I went about this the right way. Please point me in the right direction so that I could find the solution by myself.

  • 3
    $\begingroup$ $ln(3^x-3^y) \neq ln(3^x) - ln(3^y)$. $\endgroup$
    – dh16
    Sep 20, 2015 at 21:09
  • 2
    $\begingroup$ Instead from $x+y=4$, solve for either $x$ or $y$ and substitute into the second equation. $\endgroup$
    – dh16
    Sep 20, 2015 at 21:09


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