Question

question 13 (mandatory) (1 point) the growth in population of a town since 2000 is given, in thousands, by the function (p(x)=36.5(1.06)^x). in which year will the population expect to reach 70000? a) 2011 b) 2013 c) 2010 d) 2008

Explanation:

Step1: Set up the equation

Since the population is given in thousands, set $P(x)=70$ and the function is $P(x) = 36.5(1.06)^x$. So, $70=36.5(1.06)^x$.

Step2: Isolate the exponential term

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

Step3: Take the natural - logarithm of both sides

$\ln(1.9178)=\ln((1.06)^x)$. Using the property $\ln(a^b)=b\ln(a)$, 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 $x=\frac{0.640}{0.0583}\approx11$.
Since $x$ represents the number of years since 2000, the year is $2000 + 11=2011$.

Answer:

a) 2011