# How to simplify 3^((2\log_335)) How do I simplify this? This is what I have done so far: 2log_3 5=log_3 5^2=log_3(25) 3^(log_3(25))

How to simplify ${3}^{\left(2{\mathrm{log}}_{3}35\right)}$
How do I simplify this? This is what I have done so far:
$2{\mathrm{log}}_{3}5={\mathrm{log}}_{3}{5}^{2}={\mathrm{log}}_{3}\left(25\right)$
${3}^{{\mathrm{log}}_{3}\left(25\right)}$
What do I do from here? And the answer is one of these mixed solutions:
$0$
$-2$
$-\frac{\pi }{4}$
$\frac{1}{x+2}$
$±\frac{4}{25}$
$25$
$30°$
$2$
$3$
$5$
$\pi$
$\frac{\pi }{3}$
$\left(-\mathrm{\infty },2\right)$
$4\left(x+1{\right)}^{2}+3$
$-\frac{\sqrt{2}}{2}$
$-\frac{\sqrt{3}}{2}$
$\frac{\sqrt{2}}{2}$
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$m\mathrm{log}a=\mathrm{log}{a}^{m}$ when both the logarithms remain defined
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{3}^{\left(2{\mathrm{log}}_{3}5\right)}={3}^{{\mathrm{log}}_{3}\left({5}^{2}\right)}$
Now ${a}^{{\mathrm{log}}_{a}b}=b$ when the logarithm remains defined
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