# Exercise ID205

Algebra → Exponential equations and inequalities → Exponential equations
[Level: ] [Number of helps: 0] [Number of pictures: 0] [Number of steps: 10] [Number of characters: 0]

Solve the following equation.

${2}^{x-1}-{2}^{x-3}={3}^{x-2}-{3}^{x-3}$

${2}^{x-1}-{2}^{x-3}={3}^{x-2}-{3}^{x-3}\phantom{\rule{0ex}{0ex}}$ ${2}^{x}·{2}^{-1}-{2}^{x}·{2}^{-3}={3}^{x}·{3}^{-2}-{3}^{x}·{3}^{-3}\phantom{\rule{0ex}{0ex}}$ $\frac{{2}^{x}}{2}-\frac{{2}^{x}}{{2}^{3}}=\frac{{3}^{x}}{{3}^{2}}-\frac{{3}^{x}}{{3}^{3}}\phantom{\rule{0ex}{0ex}}$ ${2}^{x}\left(\frac{1}{2}-\frac{1}{{2}^{3}}\right)={3}^{x}\left(\frac{1}{{3}^{2}}-\frac{1}{{3}^{3}}\right)\phantom{\rule{0ex}{0ex}}$ ${2}^{x}\left(\frac{{2}^{2}-1}{{2}^{3}}\right)={3}^{x}\left(\frac{3-1}{{3}^{3}}\right)\phantom{\rule{0ex}{0ex}}$ ${2}^{x}\left(\frac{3}{{2}^{3}}\right)={3}^{x}\left(\frac{2}{{3}^{3}}\right)\phantom{\rule{0ex}{0ex}}$ $\frac{{2}^{x}}{{3}^{x}}=\frac{2·{2}^{3}}{{3}^{3}·3}\phantom{\rule{0ex}{0ex}}$ $\frac{{2}^{x}}{{3}^{x}}=\frac{{2}^{4}}{{3}^{4}}\phantom{\rule{0ex}{0ex}}$ ${\left(\frac{2}{3}\right)}^{x}={\left(\frac{2}{3}\right)}^{4}\phantom{\rule{0ex}{0ex}}$ $x=4\phantom{\rule{0ex}{0ex}}$
$x=4$

${a}^{n}·{a}^{m}={a}^{n+m}$

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