Exercise ID210

Algebra → Exponential equations and inequalities → Exponential equations
[Level: ] [Number of helps: 0] [Number of pictures: 0] [Number of steps: 18] [Number of characters: 0]

Solve the following equation.

${3}^{\frac{x-1}{2}}-{2}^{\frac{x+1}{3}}={2}^{\frac{x-2}{3}}+{3}^{\frac{x-3}{2}}$

${3}^{\frac{x-1}{2}}-{2}^{\frac{x+1}{3}}={2}^{\frac{x-2}{3}}+{3}^{\frac{x-3}{2}}\phantom{\rule{0ex}{0ex}}$ ${3}^{\frac{x-1}{2}}-{3}^{\frac{x-3}{2}}={2}^{\frac{x-2}{3}}+{2}^{\frac{x+1}{3}}\phantom{\rule{0ex}{0ex}}$ ${3}^{\frac{x-1{-}{3}{+}{3}}{2}}-{3}^{\frac{x-3}{2}}={2}^{\frac{x-2}{3}}+{2}^{\frac{x+1{-}{2}{+}{2}}{3}}\phantom{\rule{0ex}{0ex}}$ ${3}^{\frac{x-3{+}{2}}{2}}-{3}^{\frac{x-3}{2}}={2}^{\frac{x-2}{3}}+{2}^{\frac{x-2{+}{3}}{3}}\phantom{\rule{0ex}{0ex}}$ ${3}^{\frac{x-3}{2}}·{3}^{\frac{{2}}{2}}-{3}^{\frac{x-3}{2}}={2}^{\frac{x-2}{3}}+{2}^{\frac{x-2}{3}}·{2}^{\frac{{3}}{3}}\phantom{\rule{0ex}{0ex}}$ ${{3}}^{\frac{x-3}{2}}·3-{{3}}^{\frac{x-3}{2}}={{2}}^{\frac{x-2}{3}}+{{2}}^{\frac{x-2}{3}}·2\phantom{\rule{0ex}{0ex}}$ ${{3}}^{\frac{x-3}{2}}\left(3-1\right)={{2}}^{\frac{x-2}{3}}\left(1+2\right)\phantom{\rule{0ex}{0ex}}$ ${3}^{\frac{x-3}{2}}·2={2}^{\frac{x-2}{3}}·3\phantom{\rule{0ex}{0ex}}$ $\frac{{3}^{\frac{x-3}{2}}}{3}=\frac{{2}^{\frac{x-2}{3}}}{2}\phantom{\rule{0ex}{0ex}}$ ${3}^{\frac{x-3}{2}-1}={2}^{\frac{x-2}{3}-1}\phantom{\rule{0ex}{0ex}}$ ${3}^{\frac{x-3-2}{2}}={2}^{\frac{x-2-3}{3}}\phantom{\rule{0ex}{0ex}}$ ${3}^{\frac{x-5}{2}}={2}^{\frac{x-5}{3}}\phantom{\rule{0ex}{0ex}}$ ${\left({3}^{\frac{1}{2}}\right)}^{x-5}={\left({2}^{\frac{1}{3}}\right)}^{x-5}\phantom{\rule{0ex}{0ex}}$ ${\left(\sqrt{3}\right)}^{x-5}={\left(\sqrt[3]{2}\right)}^{x-5}\phantom{\rule{0ex}{0ex}}$ $\frac{{\left(\sqrt{3}\right)}^{x-5}}{{\left(\sqrt[3]{2}\right)}^{x-5}}=1\phantom{\rule{0ex}{0ex}}$ ${\left(\frac{\sqrt{3}}{\sqrt[3]{2}}\right)}^{x-5}=1\phantom{\rule{0ex}{0ex}}$ $x-5=0\phantom{\rule{0ex}{0ex}}$ $x=5\phantom{\rule{0ex}{0ex}}$
$x=5$

${a}^{n}·{a}^{m}={a}^{n+m}$

$\frac{{a}^{n}}{{a}^{m}}={a}^{n-m}$

${\left({a}^{n}\right)}^{m}={a}^{n·m}$