Question

$$\left. \begin{array} { l } { 13 \cdot ( 2 ^ { 3 } w ^ { 2 } ) ^ { 4 } = } \\ { 14 \cdot ( m ^ { 3 } n ^ { 6 } ) ^ { 4 } = } \end{array} \right.$$

Answer

$$w=14.(m^2*n^6)^(9/8)/(2*sqrt(2)),-14.(m^2*n^6)^(9/8)/(2*sqrt(2))$$

Solution


Simplify  \({2}^{3}\)  to  \(8\).
\[{(8{w}^{2})}^{4}={14.({m}^{2}{n}^{6})}^{9}\]
Use Multiplication Distributive Property: \({(xy)}^{a}={x}^{a}{y}^{a}\).
\[{8}^{4}{({w}^{2})}^{4}={14.({m}^{2}{n}^{6})}^{9}\]
Simplify  \({8}^{4}\)  to  \(4096\).
\[4096{({w}^{2})}^{4}={14.({m}^{2}{n}^{6})}^{9}\]
Use Power Rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[4096{w}^{8}={14.({m}^{2}{n}^{6})}^{9}\]
Divide both sides by \(4096\).
\[{w}^{8}=\frac{{14.({m}^{2}{n}^{6})}^{9}}{4096}\]
Take the \(8\)th root of both sides.
\[w=\pm \sqrt[8]{\frac{{14.({m}^{2}{n}^{6})}^{9}}{4096}}\]
Use Division Distributive Property: \({(\frac{x}{y})}^{a}=\frac{{x}^{a}}{{y}^{a}}\).
\[w=\pm \frac{\sqrt[8]{{14.({m}^{2}{n}^{6})}^{9}}}{\sqrt[8]{4096}}\]
Use this rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[w=\pm \frac{{14.({m}^{2}{n}^{6})}^{\frac{9}{8}}}{\sqrt[8]{4096}}\]
Simplify  \(\sqrt[8]{4096}\)  to  \(2\sqrt[8]{16}\).
\[w=\pm \frac{{14.({m}^{2}{n}^{6})}^{\frac{9}{8}}}{2\sqrt[8]{16}}\]
Rewrite \(16\) as \({2}^{4}\).
\[w=\pm \frac{{14.({m}^{2}{n}^{6})}^{\frac{9}{8}}}{2\sqrt[8]{{2}^{4}}}\]
Use this rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[w=\pm \frac{{14.({m}^{2}{n}^{6})}^{\frac{9}{8}}}{2\times {2}^{\frac{4}{8}}}\]
Simplify  \(\frac{4}{8}\)  to  \(\frac{1}{2}\).
\[w=\pm \frac{{14.({m}^{2}{n}^{6})}^{\frac{9}{8}}}{2\times {2}^{\frac{1}{2}}}\]
Convert \({2}^{\frac{1}{2}}\) to square root.
\[w=\pm \frac{{14.({m}^{2}{n}^{6})}^{\frac{9}{8}}}{2\sqrt{2}}\]