## College Algebra (6th Edition)

118 ft The point (118,1) is on the graph of $f(x)$.
$f(x)=20(0.975)^{x}$ We want $f(x)=1$ (percent). Insert and solve for x: $1=20(0.975)^{x}\quad$...$/\div 20$ $\displaystyle \frac{1}{20}=0.975^{x}\qquad .../$apply ln() to both sides... $\ln 0.05=x\ln 0.975\quad$...$/\div\ln 0.975$ $x=\displaystyle \frac{\ln 0.05}{\ln 0.975}\approx$118.325104425$\approx 118$ ft At about x = 118 ft depth, there is $1\%$ of surface sunlight . $\mathrm{f}(118)\approx 1$ The point (118,1) is on the graph of $f(x)$.