## Precalculus (6th Edition) Blitzer

Solution set = $\displaystyle \{\frac{e^{5}}{2}\}.$ $x\approx 74.21$
For the equation to be defined, logarithms must have positive arguments $2x \gt 0$ $x \gt 0 \qquad (*)$ This is the condition eventual solutions must satisfy. $6\ln 2x=30\qquad$... divide with $6$ $\ln 2x=5\qquad$... apply $e^{(...)}$, the inverse of $\ln$ $e^{\ln 2x}=e^{5}$ $2x=e^{5}\qquad$... divide with $2$ $x=\displaystyle \frac{e^{5}}{2}\qquad$... round to 2 decimal places $x=74.21$ (satisfies the condition (*), and is thus a valid solution.)