Question

question 6 (mandatory) (1 point) the growth in population of a town since 2000 is given, in thousands, by the function $p(n)=36.5(1.06)^{n}$. in which year will the population expect to reach 70 000? a) 2008 b) 2010 c) 2013 d) 2011

Explanation:

Step1: Set up the equation

Since the population function $P(x)=36.5(1.06)^{x}$ is in thousands and we want the population to be 70000 (or 70 in thousands), we set up the equation $70 = 36.5(1.06)^{x}$.

Step2: Isolate the exponential term

Divide both sides of the equation by 36.5: $\frac{70}{36.5}=(1.06)^{x}$, so $1.9178=(1.06)^{x}$.

Step3: Take the natural - logarithm of both sides

$\ln(1.9178)=\ln((1.06)^{x})$. Using the property of logarithms $\ln(a^{b}) = b\ln(a)$, we get $\ln(1.9178)=x\ln(1.06)$.

Step4: Solve for x

$x=\frac{\ln(1.9178)}{\ln(1.06)}$. Calculate $\ln(1.9178)\approx0.640$ and $\ln(1.06)\approx0.0583$. Then $x=\frac{0.640}{0.0583}\approx11$.

Answer:

Since $x$ represents the number of years since 2000, the year is $2000 + 11=2011$. So the answer is d) 2011.