## Precalculus (6th Edition) Blitzer

$277.2589$
Re-arrange as: $e^{0.005x}=4$ We need to take $\ln$ of each side of the equation. $\ln e^{0.005x} =\ln 4$ Simplify it by applying rule $\log{a^n}=n \log x$ to obtain: $0.005x= \ln 1.4$ This implies that $x = \dfrac{\ln 4}{0.005} \approx 277.2589$