# Exercise ID329

Algebra → Logarithm → Logarithm equations
[Level: ] [Number of helps: 0] [Number of pictures: 0] [Number of steps: 22] [Number of characters: 87]

Evaluate the following expression:

${x}^{\frac{{\mathrm{log}}_{10}\left(x+5\right)}{3}}={10}^{5+{\mathrm{log}}_{10}\left(x\right)}\phantom{\rule{0ex}{0ex}}$

${x}^{\frac{{\mathrm{log}}_{10}x+5}{3}}=\frac{{10}^{5+{\mathrm{log}}_{10}x}}{{l}{o}{{g}}_{10}{10}}\phantom{\rule{0ex}{0ex}}$ ${x}^{\frac{1}{3}\left({\mathrm{log}}_{10}x+5\right)}=\frac{{10}^{5+{\mathrm{log}}_{10}x}}{{1}}\phantom{\rule{0ex}{0ex}}$ ${l}{o}{{g}}_{10}{x}^{\frac{1}{3}{·}\left({\mathrm{log}}_{10}x+5\right)}={l}{o}{{g}}_{10}{10}^{{5}{+}{l}{o}{{g}}_{10}{x}}\phantom{\rule{0ex}{0ex}}$ $\frac{1}{3}{·}\left({\mathrm{log}}_{10}x+5\right)·{l}{o}{{g}}_{10}{x}=\left(5+lo{g}_{10}x\right)·{l}{o}{{g}}_{10}{10}\phantom{\rule{0ex}{0ex}}$ $\frac{1}{3}·\left({\mathrm{log}}_{10}x+5\right)·{\mathrm{log}}_{10}x=\left(5+{\mathrm{log}}_{10}x\right)·1\phantom{\rule{0ex}{0ex}}$ ${\mathrm{log}}_{10}x=t\phantom{\rule{0ex}{0ex}}$ $\frac{1}{3}·\left(t+5\right)·t=5+t\phantom{\rule{0ex}{0ex}}$ ${t}_{1,2}=\frac{-2±\sqrt{{2}^{2}-4·1·\left(-15\right)}}{2}\phantom{\rule{0ex}{0ex}}$ ${t}_{1,2}=\frac{-2±\sqrt{4+60}}{2}=\frac{-2±\sqrt{64}}{2}\phantom{\rule{0ex}{0ex}}$ ${t}_{1,2}=\frac{-2±8}{2}\phantom{\rule{0ex}{0ex}}$ ${t}_{1,2}=-1±4\phantom{\rule{0ex}{0ex}}$ ${\mathrm{log}}_{10}{x}_{1}=3\phantom{\rule{0ex}{0ex}}$ ${x}_{1}={10}^{3}=1000\phantom{\rule{0ex}{0ex}}$ ${\mathrm{log}}_{10}{x}_{2}=-5\phantom{\rule{0ex}{0ex}}$ ${x}_{2}={10}^{-5}=\frac{1}{{10}^{5}}\phantom{\rule{0ex}{0ex}}$ $x\in \left\{{10}^{3},{10}^{-5}\right\}\phantom{\rule{0ex}{0ex}}$