Solve (e^x + e^-x) ÷ e^x – e^-x= 2

\(\displaystyle \L\begin{array}{rcl}
\frac{{e^x + e^{ - x} }}{{e^x - e^{ - x} }} & = & 2\quad ,x \not= 0 \\
e^x + e^{ - x} & = & 2e^x - 2e^{ - x} \\
e^x - 3e^{ - x} & = & 0 \\
e^{2x} & = & 3 \\
x & = & \ln \left( {\sqrt 3 } \right) \\
\end{array}\)
 
Top