Chapter 5 - 5.5 - Exponential and Logarithmic Models - 5.5 Exercises - Page 405: 9

$8.9438$ and $1834.346$

We have $A=Pe^{rt} \implies 2P=Pe^{rt}$ $\implies 2 =e^{rt}$ Take the $\log$ on each side. $\log 2 = \log e^{rt} \implies t = \dfrac{\ln 2}{r}=\dfrac{\ln 2}{0.0775} \approx 8.9438$ Now, $A=Pe^{rt} = 750 \times e^{8.9438 \times 10}$ or, $A \approx 1834.346$ Our results are: $8.9438$ and $1834.346$