81^(1/(log _5 3)) +27^((log _9 36)) + 3^(1/(log _7 9))?

Aug 29, 2015

${81}^{\frac{1}{{\log}_{5} 3}} + {27}^{{\log}_{9} 36} + {3}^{\frac{1}{{\log}_{7} 9}} = 841 + \sqrt{7}$

Explanation:

$\frac{1}{{\log}_{5} 3} = {\log}_{3} \left(5\right)$

So:

${81}^{\frac{1}{{\log}_{5} 3}} = {\left({3}^{4}\right)}^{{\log}_{3} \left(5\right)} = {3}^{4 {\log}_{3} \left(5\right)} = {\left({3}^{{\log}_{3} \left(5\right)}\right)}^{4} = {5}^{4} = 625$

${\log}_{9} 36 = \frac{{\log}_{3} \left(36\right)}{{\log}_{3} \left(9\right)} = {\log}_{3} \frac{36}{{\log}_{3} \left({3}^{2}\right)} = {\log}_{3} \frac{36}{2}$

So:

${27}^{{\log}_{9} 36} = {\left({3}^{3}\right)}^{{\log}_{3} \frac{36}{2}} = {\left({3}^{{\log}_{3} \left(36\right)}\right)}^{\frac{3}{2}} = {36}^{\frac{3}{2}} = {\left({6}^{2}\right)}^{\frac{3}{2}} = {6}^{3} = 216$

$\frac{1}{{\log}_{7} 9} = {\log}_{9} \left(7\right) = {\log}_{3} \frac{7}{2}$

So:

${3}^{\frac{1}{{\log}_{7} 9}} = {3}^{{\log}_{3} \frac{7}{2}} = {\left({3}^{{\log}_{3} \left(7\right)}\right)}^{\frac{1}{2}} = {7}^{\frac{1}{2}} = \sqrt{7}$

Putting this all together:

${81}^{\frac{1}{{\log}_{5} 3}} + {27}^{{\log}_{9} 36} + {3}^{\frac{1}{{\log}_{7} 9}} = 625 + 216 + \sqrt{7} = 841 + \sqrt{7}$