# Exercise ID324

Algebra → Logarithm → Logarithm equations
[Level: ] [Number of helps: 0] [Number of pictures: 0] [Number of steps: 7] [Number of characters: 0]

Evaluate the following expression:

${36}^{{\mathrm{log}}_{6}\left(5\right)}+{10}^{1-{\mathrm{log}}_{10}\left(2\right)}-{3}^{{\mathrm{log}}_{9}\left(36\right)}=\phantom{\rule{0ex}{0ex}}$

${36}^{{\mathrm{log}}_{6}\left(5\right)}+{10}^{1}×{10}^{{\mathrm{log}}_{10}\left({2}^{-1}\right)}-{3}^{{\mathrm{log}}_{9}\left(36\right)}=\phantom{\rule{0ex}{0ex}}$ ${6}^{2{\mathrm{log}}_{6}\left(5\right)}+{10}^{1}×{10}^{{\mathrm{log}}_{10}\left({2}^{-1}\right)}-{3}^{{\mathrm{log}}_{{3}^{2}}\left(36\right)}=\phantom{\rule{0ex}{0ex}}$ ${6}^{{\mathrm{log}}_{6}\left({5}^{2}\right)}+10×{2}^{-1}-{3}^{\frac{1}{2}{\mathrm{log}}_{3}\left(36\right)}=\phantom{\rule{0ex}{0ex}}$ $25+10×{2}^{-1}-{3}^{{\mathrm{log}}_{3}\left(\sqrt{36}\right)}=\phantom{\rule{0ex}{0ex}}$ $25+10×{2}^{-1}-\left(\sqrt{36}\right)=\phantom{\rule{0ex}{0ex}}$ $25+5-6=\phantom{\rule{0ex}{0ex}}$ $24\phantom{\rule{0ex}{0ex}}$
$24\phantom{\rule{0ex}{0ex}}$