Question

Please help!!! will mark brainliest. only 2 questions, and worth 50 pts.

Answer 1

Answer:

1)  13,434

2)  During 2038 (by 2039)

Step-by-step explanation:

We can model this problem using an exponential equation.

General form of an exponential equation: $y=ab^x$

where:

• a is the initial value
• b is the base (or growth factor) in decimal form
• x is the independent variable
• y is the dependent variable

If b > 1 then it is an increasing function

If 0 < b < 1 then it is a decreasing function

As the population increases by 3% each year, each year it will be 103% of the previous year.  Therefore, the growth factor in decimal form is 1.03

Given:

• a = 8,888
• b = 1.03
• x = time (in years)
• y = population

Substituting the given values into the equation:

$y=8888(1.03)^x$

To find the population in 2025:

2025 - 2011 = 14

Therefore, set x = 14 and solve for y:

$\begin{aligned}\implies y &=8888(1.03)^{14}\\& =13433.89747\\& = 13,434\end{aligned}$

To find the year in which the population will reach 20000, set y = 20000 and solve for x:

$\implies 8888(1.03)^x=20000$

$\implies (1.03)^x=\dfrac{20000}{8888}$

$\implies \ln (1.03)^x=\ln \left(\dfrac{20000}{8888}\right)$

$\implies x\ln (1.03)=\ln \left(\dfrac{20000}{8888}\right)$

$\implies x=\dfrac{\ln \left(\frac{20000}{8888}\right)}{\ln (1.03)}$

$\implies x=27.43785809$

2011 + 27.437... = 2038.437...

Therefore, the population will reach 20,000 some time during 2038.