# Can someone help me out with this question about logs? please 3^(2x)−2^(2y)=17 Find x+y. Here is what I did so far: Let m=3^(2x) and let n=2^(2y) x=(log_3m)/(2), y=(log_2n)/2

${3}^{2x}-{2}^{2y}=17$
Find $x+y$
Here is what I did so far: Let $m={3}^{2x}$ and let $n={2}^{2y}$, $x=\frac{{\mathrm{log}}_{3}m}{2}$, $y=\frac{{\mathrm{log}}_{2}n}{2}$
$x+y=\frac{{\mathrm{log}}_{3}m+{\mathrm{log}}_{2}n}{2}$
x+y= (base(3)17+n)+(base(2)n)/2
don't know what to do from there
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Miah Scott
$17={3}^{2x}-{2}^{2y}=\left({3}^{x}-{2}^{y}\right)\left({3}^{x}+{2}^{y}\right)$
and now remember $\phantom{\rule{thickmathspace}{0ex}}17\phantom{\rule{thickmathspace}{0ex}}$ is prime. I don't think logarithms are required here.
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