Mathematic Problems
$\frac{\left({27}^{\frac{1}{{\mathrm{log}}_{2}\left(3\right)}}+{5}^{{\mathrm{log}}_{25}\left(49\right)}\right)×\left({81}^{\frac{1}{{\mathrm{log}}_{4}\left(9\right)}}-{8}^{{\mathrm{log}}_{4}\left(9\right)}\right)}{3+{5}^{\frac{1}{{\mathrm{log}}_{16}\left(25\right)}}×{5}^{{\mathrm{log}}_{5}\left(9\right)}}=\phantom{\rule{0ex}{0ex}}$
$\frac{\left({27}^{{\mathrm{log}}_{3}\left(2\right)}+{5}^{\frac{1}{2}×2{\mathrm{log}}_{5}\left(7\right)}\right)×\left({81}^{{\mathrm{log}}_{9}\left(4\right)}-{8}^{{\mathrm{log}}_{4}\left(9\right)}\right)}{3+{5}^{\frac{1}{2}×2{\mathrm{log}}_{5}\left(4\right)}×{5}^{{\mathrm{log}}_{5}\left(3\right)}}=\phantom{\rule{0ex}{0ex}}$
$\frac{\left({3}^{3{\mathrm{log}}_{3}\left(2\right)}+{5}^{{\mathrm{log}}_{5}\left(7\right)}\right)×\left({3}^{4{\mathrm{log}}_{9}\left(4\right)}-{2}^{3{\mathrm{log}}_{4}\left(9\right)}\right)}{3+{5}^{{\mathrm{log}}_{5}\left(4\right)}×{5}^{{\mathrm{log}}_{5}\left(3\right)}}=\phantom{\rule{0ex}{0ex}}$
$\frac{\left({3}^{{\mathrm{log}}_{3}\left({2}^{3}\right)}+{5}^{{\mathrm{log}}_{5}\left(7\right)}\right)×\left({3}^{\frac{4}{2}{\mathrm{log}}_{3}\left(4\right)}-{2}^{2×\frac{3}{2}{\mathrm{log}}_{2}\left(3\right)}\right)}{3+{5}^{{\mathrm{log}}_{5}\left(4\right)}×{5}^{{\mathrm{log}}_{5}\left(3\right)}}=\phantom{\rule{0ex}{0ex}}$
$\frac{\left({2}^{3}+7\right)×\left({3}^{{\mathrm{log}}_{3}\left({4}^{2}\right)}-{2}^{\frac{6}{2}{\mathrm{log}}_{2}\left(3\right)}\right)}{3+4×3}=\phantom{\rule{0ex}{0ex}}$
$\frac{\left(8+7\right)×\left({4}^{2}-{2}^{{\mathrm{log}}_{2}\left({3}^{3}\right)}\right)}{3+4×3}=\phantom{\rule{0ex}{0ex}}$
$\frac{\left(8+7\right)×\left(16-27\right)}{3+4×3}=\phantom{\rule{0ex}{0ex}}$
$-11\phantom{\rule{0ex}{0ex}}$
$-11\phantom{\rule{0ex}{0ex}}$