Chapter 5 - 5.5 - Exponential and Logarithmic Models - 5.5 Exercises - Page 405: 8

$6.601$ and $2143.24$

We have $A=Pe^{rt} \implies 2P=Pe^{rt}$ $\implies 2 =e^{rt}$ Take the $\log$ on each side. $\log 2 = \log e^{rt} \implies t = \dfrac{\ln 2}{r}=\dfrac{\ln 2}{0.105} \approx 6.601$ Now, $A=Pe^{rt} = 750 \times e^{0.105 \times 10}$ or, $A \approx 2143.24$ Our results are: $6.601$ and $2143.24$