## Algebra and Trigonometry 10th Edition

(a) $t=27.73~years$ (b) $t=43.94~years$
$A=Pe^{rt}~~$ (Continuous compounding) $A=2500e^{0.025t}$ (a) We want $A=5000$ $5000=2500e^{0.025t}$ $2=e^{0.025t}$ $\ln2=\ln e^{0.025t}$ $\ln2=0.025t$ $t=\frac{\ln2}{0.025}=27.73~years$ (b) We want $A=7500$ $7500=2500e^{0.025t}$ $3=e^{0.025t}$ $\ln3=\ln e^{0.025t}$ $\ln3=0.025t$ $t=\frac{\ln3}{0.025}=43.94~years$