Chapter 7 - Section 7.2 - Exponential Change and Separable Differential Equations - Exercises - Page 410: 30

$1250$

Given: $y(3)=10,000$ and $y(5)=40,000$ The exponential growth can be written as: $y=y_0e^{kt}$ ...(1) Then, $10,000=y_0 e^{3k}$ and $40,000=y_0 e^{5k}$ This implies that $y_0 e^{5k}=4y_0 e^{3k}$ or, $e^{2k}=4 \implies k= \ln 2$ Equation (1) becomes: $y=y_0e^{(\ln 2)t}$ or, $10,000=y_0e^{3 \ln 2} \implies 10,000=y_0e^{\ln 2^3}$ so, $10,000=8y_0 \implies y_0=\dfrac{10,000}{8}=1250$