Question

question 20 (mandatory) (1 point) the growth in population of a town since 2000 is given, in thousands, by the function f(x)=36.5(1.06)^x. in which year will the population expect to reach 70000? a) 2008 b) 2010 c) 2013 d) 2011

Explanation:

Step1: Set up the equation

The population function is $f(x)=36.5(1.06)^x$, and we want $f(x) = 70$ (since the population is in thousands and we want 70000). So, $36.5(1.06)^x=70$.

Step2: Isolate the exponential term

Divide both sides by 36.5: $(1.06)^x=\frac{70}{36.5}\approx1.9178$.

Step3: Take the natural - logarithm of both sides

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

Step4: Solve for x

$x = \frac{\ln(1.9178)}{\ln(1.06)}\approx\frac{0.6409}{0.0583}\approx11$. Since $x$ represents the number of years since 2000, the year is $2000 + 11=2011$.

Answer:

d) 2011