## College Algebra 7th Edition

(a) $x=\frac{1-\ln 12}{4}$ (b) $x\approx -0.371227$
(a) We need to solve: $8+e^{1-4x}=20$ $e^{1-4x}=20-8=12$ We take the log of both sides and use log rules to simplify: $\ln e^{1-4x}=\ln 12$ $1-4x=\ln 12$ $x=\frac{1-\ln 12}{4}$ (b) We solve using a calculator: $x\approx -0.371227$