Question

$$A^{\prime}nB^{>}$$

Answer

$$A>1/(r^(1/p)*IM^(1/p)*m^(1/p)*e^(1/p)*nB^(1/p))$$

Solution


Divide both sides by \(r\).
\[{A}^{p}\imath me{nB}^{}>\frac{1}{r}\]
Divide both sides by \(\imath \).
\[{A}^{p}me{nB}^{}>\frac{\frac{1}{r}}{\imath }\]
Simplify  \(\frac{\frac{1}{r}}{\imath }\)  to  \(\frac{1}{r\imath }\).
\[{A}^{p}me{nB}^{}>\frac{1}{r\imath }\]
Divide both sides by \(m\).
\[{A}^{p}e{nB}^{}>\frac{\frac{1}{r\imath }}{m}\]
Simplify  \(\frac{\frac{1}{r\imath }}{m}\)  to  \(\frac{1}{r\imath m}\).
\[{A}^{p}e{nB}^{}>\frac{1}{r\imath m}\]
Divide both sides by \(e\).
\[{A}^{p}{nB}^{}>\frac{\frac{1}{r\imath m}}{e}\]
Simplify  \(\frac{\frac{1}{r\imath m}}{e}\)  to  \(\frac{1}{r\imath me}\).
\[{A}^{p}{nB}^{}>\frac{1}{r\imath me}\]
Divide both sides by \({nB}^{}\).
\[{A}^{p}>\frac{\frac{1}{r\imath me}}{{nB}^{}}\]
Simplify  \(\frac{\frac{1}{r\imath me}}{{nB}^{}}\)  to  \(\frac{1}{r\imath me{nB}^{}}\).
\[{A}^{p}>\frac{1}{r\imath me{nB}^{}}\]
Take the \((p)\)th root of both sides.
\[A>\sqrt[p]{\frac{1}{r\imath me{nB}^{}}}\]
Use Division Distributive Property: \({(\frac{x}{y})}^{a}=\frac{{x}^{a}}{{y}^{a}}\).
\[A>\frac{1}{\sqrt[p]{r\imath me{nB}^{}}}\]
Use Multiplication Distributive Property: \({(xy)}^{a}={x}^{a}{y}^{a}\).
\[A>\frac{1}{\sqrt[p]{r}\sqrt[p]{\imath }\sqrt[p]{m}\sqrt[p]{e}\sqrt[p]{{nB}^{}}}\]
Use this rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[A>\frac{1}{\sqrt[p]{r}\sqrt[p]{\imath }\sqrt[p]{m}\sqrt[p]{e}\sqrt[p]{nB}}\]