Question

NO LINKS!!!! Exponential Growth and Decay Part 2​

Answer 1

Problem 4

• a = 10800 = initial population
• b = 1 + r = 1 + (-0.025) = 0.975 is the decay factor

The template of
`y = a*b^x` becomes
`y = 10800*0.975^x` to represent the exponential function.

• x = number of years since 2002
• y = population

We want to know when the population reaches half of 10800, so we want to know when the population is 10800/2 = 5400

Plug in y = 5400 and solve for x.

`y = 10800*0.975^x\\\\5400 = 10800*0.975^x\\\\0.975^x = 5400/10800\\\\0.975^x = 0.5\\\\\log(0.975^x) = \log(0.5)\\\\x\log(0.975) = \log(0.5)\\\\x = \log(0.5)/\log(0.975)\\\\x \approx 27.377851\\\\x \approx 28\\\\`

I rounded up to the nearest whole number because x = 27 leads to y = 5452, which is not 5400 or smaller.

Luckily, x = 28 leads to y = 5315 which gets over the hurdle of being 5400 or smaller.

Add 28 years onto the starting year 2002 and we get to 2002+28 = 2030

The population reaches half of its original amount in the year 2030.

Answers:

• The exponential function is
  `y = 10800*0.975^x`
• It takes 28 years to get to half the population. This occurs in the year 2030

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Problem 5

• a = 28750 = starting value for the car
• b = 1 + r = 1 + (-0.12) = 0.88 = decay factor

If the car loses 12% of its value each year, then it keeps the remaining 88%

Plug those values into
`y = a*b^x`.

We find the equation is
`y = 28750*0.88^x` where,

• x = number of years since 2012
• y = car's value

Replace y with 10,000 and solve for x.

`y = 28750*0.88^x\\\\10000 = 28750*0.88^x\\\\0.88^x = 10000/28750\\\\0.88^x \approx 0.347826\\\\\log(0.88^x) \approx \log(0.347826)\\\\x\log(0.88) \approx \log(0.347826)\\\\x \approx \log(0.347826)/\log(0.88)\\\\x \approx 8.261168\\\\x \approx 9\\\\`

Like in the previous problem, we round up so we clear the hurdle.

Adding 9 years onto 2012 gets us to 2012+9 = 2021

Answers:

• The function is
  `y = 28750*0.88^x`
• It takes about 9 years, and it occurs in the year 2021

Answer 2

Exponential Function

General form of an exponential function:
`y=ab^x`

where:

• a is the initial value (y-intercept)
• b is the base (growth/decay 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

Question 4

Given:

• a = 10,800
• b = decrease of 2.5% = 0.975
• x = time (in years)
• y = population

As the population is decreasing by 2.5% each year, the population will be 100% - 2.5% = 97.5% of the previous year. Therefore, the base is 0.975.

Final equation:
`\large \text{$ y=10800(0.975)^x $}`

Half of population: 10800 ÷ 2 = 5400

`\large \begin{aligned}y & =5400\\\implies 10800(0.975)^x & =5400\\(0.975)^x & = (5400)/(10800)\\(0.975)^x & = 0.5\\\ln (0.975)^x & = \ln 0.5\\x \ln 0.975 & = \ln 0.5\\x & = (\ln 0.5)/(\ln 0.975)\\x & = 27.377785123\end{aligned}`

2002 + 27.37785... = 2029.37785...

Therefore, the population will reach half during 2029 (by 2030).

Question 5

Given:

• a = 28,750
• b = decrease of 12% = 0.88
• x = time (in years)
• y = value (in dollars)

As the value is decreasing by 12% each year, the value will be 100% - 12% = 88% of the previous year. Therefore, the base is 0.88.

Final equation:
`\large \text {$ y=28750(0.88)^x $}`

Find when the car is worth $10,000:

`\large \begin{aligned}y & = 10000\\\implies 28750(0.88)^x & = 10000\\(0.88)^x & = (10000)/(28750)\\(0.88)^x & = (8)/(23)\\\ln (0.88)^x & =\ln \left((8)/(23)\right)\\x \ln (0.88) & =\ln \left((8)/(23)\right)\\x & =(\ln \left((8)/(23)\right))/(\ln (0.88))\\x & = 8.26116578\end{aligned}`

2012 + 8.26116578.. = 2020.26116578..

Therefore, the value of the car will reach $10,000 during 2020 (by 2021).