Question

$$\epsilon x_{\eta-n}x=n$$

Answer

$$p=(1+x)/(e^2*IM*s*l*o*x*t*a)$$

Solution


Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[{e}^{2}ps\imath lonxta-nx=n\]
Regroup terms.
\[{e}^{2}\imath pslonxta-nx=n\]
Regroup terms.
\[-nx+{e}^{2}\imath pslonxta=n\]
Add \(nx\) to both sides.
\[{e}^{2}\imath pslonxta=n+nx\]
Factor out the common term \(n\).
\[{e}^{2}\imath pslonxta=n(1+x)\]
Cancel \(n\) on both sides.
\[{e}^{2}\imath psloxta=1+x\]
Divide both sides by \({e}^{2}\).
\[\imath psloxta=\frac{1+x}{{e}^{2}}\]
Divide both sides by \(\imath \).
\[psloxta=\frac{\frac{1+x}{{e}^{2}}}{\imath }\]
Simplify  \(\frac{\frac{1+x}{{e}^{2}}}{\imath }\)  to  \(\frac{1+x}{{e}^{2}\imath }\).
\[psloxta=\frac{1+x}{{e}^{2}\imath }\]
Divide both sides by \(s\).
\[ploxta=\frac{\frac{1+x}{{e}^{2}\imath }}{s}\]
Simplify  \(\frac{\frac{1+x}{{e}^{2}\imath }}{s}\)  to  \(\frac{1+x}{{e}^{2}\imath s}\).
\[ploxta=\frac{1+x}{{e}^{2}\imath s}\]
Divide both sides by \(l\).
\[poxta=\frac{\frac{1+x}{{e}^{2}\imath s}}{l}\]
Simplify  \(\frac{\frac{1+x}{{e}^{2}\imath s}}{l}\)  to  \(\frac{1+x}{{e}^{2}\imath sl}\).
\[poxta=\frac{1+x}{{e}^{2}\imath sl}\]
Divide both sides by \(o\).
\[pxta=\frac{\frac{1+x}{{e}^{2}\imath sl}}{o}\]
Simplify  \(\frac{\frac{1+x}{{e}^{2}\imath sl}}{o}\)  to  \(\frac{1+x}{{e}^{2}\imath slo}\).
\[pxta=\frac{1+x}{{e}^{2}\imath slo}\]
Divide both sides by \(x\).
\[pta=\frac{\frac{1+x}{{e}^{2}\imath slo}}{x}\]
Simplify  \(\frac{\frac{1+x}{{e}^{2}\imath slo}}{x}\)  to  \(\frac{1+x}{{e}^{2}\imath slox}\).
\[pta=\frac{1+x}{{e}^{2}\imath slox}\]
Divide both sides by \(t\).
\[pa=\frac{\frac{1+x}{{e}^{2}\imath slox}}{t}\]
Simplify  \(\frac{\frac{1+x}{{e}^{2}\imath slox}}{t}\)  to  \(\frac{1+x}{{e}^{2}\imath sloxt}\).
\[pa=\frac{1+x}{{e}^{2}\imath sloxt}\]
Divide both sides by \(a\).
\[p=\frac{\frac{1+x}{{e}^{2}\imath sloxt}}{a}\]
Simplify  \(\frac{\frac{1+x}{{e}^{2}\imath sloxt}}{a}\)  to  \(\frac{1+x}{{e}^{2}\imath sloxta}\).
\[p=\frac{1+x}{{e}^{2}\imath sloxta}\]