Question

$$(0^{\underline{3}}50^{2}+80-y)y=e^{2x}$$

Answer

x=ln(y*(80-y))/2

Solution


Simplify  \({0}^{3}\)  to  \(0\).
\[(0\times {50}^{2}+80-y)y={e}^{2x}\]
Simplify  \({50}^{2}\)  to  \(2500\).
\[(0\times 2500+80-y)y={e}^{2x}\]
Simplify  \(0\times 2500\)  to  \(0\).
\[(0+80-y)y={e}^{2x}\]
Simplify  \(0+80-y\)  to  \(80-y\).
\[(80-y)y={e}^{2x}\]
Regroup terms.
\[y(80-y)={e}^{2x}\]
Use Definition of Natural Logarithm: \({e}^{y}=x\) if and only if \(\ln{x}=y\).
\[\ln{(y(80-y))}=2x\]
Divide both sides by \(2\).
\[\frac{\ln{(y(80-y))}}{2}=x\]
Switch sides.
\[x=\frac{\ln{(y(80-y))}}{2}\]