Solve for x: 2e^{2x}-e^{x}=0
Let e^{x}=u
2u^{2}-u=0
u(2u-1)=0
u=0 or \frac{1}{2}
e^{x}=0 OR e^{x}=\frac{1}{2}
x=\ln 0 OR x=\ln \frac{1}{2}
ln 0 is an invalid solution, does not exist.
\therefore x=\ln \frac{1}{2}
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