Question

the exponential model $a = 25.4e^{0.02t}$ describes the population, $a$, of a country in millions, $t$ years after 2003. use the model to determine when the population of the country will be 38 million. the population of the country will be 38 million in \boxed{}. (round to the nearest year as needed.)

Explanation:

Step1: Substitute \(A = 38\) into the model

Given \(A=25.4e^{0.02t}\), when \(A = 38\), we have the equation \(38=25.4e^{0.02t}\).
First, divide both sides by \(25.4\): \(\frac{38}{25.4}=e^{0.02t}\). So, \(e^{0.02t}\approx1.5\).

Step2: Take the natural - logarithm of both sides

Using the property \(\ln(e^{x})=x\), if \(y = e^{x}\), then \(x=\ln(y)\).
For \(e^{0.02t}=1.5\), we take the natural logarithm of both sides: \(\ln(e^{0.02t})=\ln(1.5)\).
Since \(\ln(e^{0.02t}) = 0.02t\), the equation becomes \(0.02t=\ln(1.5)\).
We know that \(\ln(1.5)\approx0.4055\).

Step3: Solve for \(t\)

From \(0.02t=\ln(1.5)\approx0.4055\), we can solve for \(t\) by dividing both sides by \(0.02\).
\(t=\frac{\ln(1.5)}{0.02}\approx\frac{0.4055}{0.02}=20.275\approx20\)

Answer:

The population of the country will be 38 million in \(2023\) (since \(t\approx20\) years after 2003).