happyvilles population can be modeled using the function p(t)=24000e^0.21t, where p(t) is the population t…

Explanation:

Step1: Set up the equation for part a

Set $P(t)=48000$ in $P(t) = 24000e^{0.21t}$, so $48000=24000e^{0.21t}$.

Step2: Simplify the equation

Divide both sides by 24000: $\frac{48000}{24000}=e^{0.21t}$, which gives $2 = e^{0.21t}$.

Step3: Take the natural - logarithm of both sides

$\ln(2)=\ln(e^{0.21t})$. Since $\ln(e^{x}) = x$, we have $\ln(2)=0.21t$.

Step4: Solve for $t$

$t=\frac{\ln(2)}{0.21}\approx\frac{0.6931}{0.21}\approx 3.3$. Rounding to the nearest year, $t = 3$. The year is $2011 + 3=2014$.

Step5: Set up the equation for part b

Set $P(t)=95000$ in $P(t)=24000e^{0.21t}$, so $95000 = 24000e^{0.21t}$.

Step6: Simplify the equation

Divide both sides by 24000: $\frac{95000}{24000}=e^{0.21t}$, which gives $\frac{95}{24}=e^{0.21t}$.

Step7: Take the natural - logarithm of both sides

$\ln(\frac{95}{24})=\ln(e^{0.21t})$. Since $\ln(e^{x}) = x$, we have $\ln(\frac{95}{24})=0.21t$.

Step8: Solve for $t$

$t=\frac{\ln(\frac{95}{24})}{0.21}\approx\frac{\ln(3.9583)}{0.21}\approx\frac{1.376}{0.21}\approx 6.55$. Rounding to the nearest year, $t = 7$. The year is $2011+7 = 2018$.

Answer:

a) 2014 b) 2018