## College Algebra 7th Edition

(a) $x=\frac{\ln 45}{\ln 3}-1$ (b) $x\approx 2.464974$
(a) We need to solve: $2(5+3^{x+1})=100$ $5+3^{x+1}=\frac{100}{2}=50$ $3^{x+1}=50-5=45$ We take the log of both sides and use log rules to simplify (we choose the natural log, but any other base would work as well): $\ln 3^{x+1}=\ln 45$ $(x+1) \ln3 = \ln 45$ $x+1=\frac{\ln 45}{\ln 3}$ $x=\frac{\ln 45}{\ln 3}-1$ (b) We solve using a calculator: $x\approx 2.464974$