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 is given in thousands, set $P(n)=70$ and the function is $P(n) = 36.5(1.06)^{n}$. So, $70=36.5(1.06)^{n}$.

Step2: Isolate the exponential term

Divide both sides by 36.5: $\frac{70}{36.5}=(1.06)^{n}$, which simplifies to approximately $1.9178=(1.06)^{n}$.

Step3: Take the natural - logarithm of both sides

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

Step4: Solve for n

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

Answer:

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