Question

NO LINKS!!!! Exponential Growth and Decay Part 3​

Answer 1

Compound interest formula

`\Large \text{$ \sf A=P(1+(r)/(n))^(nt) $}`

where:

• A = final amount
• P = principal amount
• r = interest rate (in decimal form)
• n = number of times interest applied per time period
• t = number of time periods elapsed

Question 6

Given:

• A = $35,000
• P = $8,000
• r = 10% = 0.1
• n = 4
• t = time (in years)

`\begin{aligned}8000\left(1+(0.1)/(4)\right)^(4t) & = 35000\\(1.025)^(4t) & = (35000)/(8000)\\(1.025)^(4t) & = 4.375}\\\ln (1.025)^(4t) & = \ln 4.375\\4t \ln 1.025 & = \ln 4.375\\4t & = (\ln 4.375)/(\ln 1.025)\\t & = 0.25\left((\ln 4.375)/(\ln 1.025)\right)\\t & = 14.9427947\end{aligned}`

Therefore, she will have $35,000 in 15 years.

Question 7

Given:

• A = $43,000
• P = $32,500
• r = 6% = 0.06
• n = 1
• t = time (in years)

`\begin{aligned}32500\left(1+(0.06)/(1)\right)^(1t) & = 43000\\32500(1.06)^(t) & = 43000\\(1.06)^(t) & = (43000)/(32500)\\(1.06)^(t) & = (86)/(65)\\\ln (1.06)^(t) & = \ln \left((86)/(65)\right)\\t \ln 1.06 & = \ln \left((86)/(65)\right)\\t & = (\ln \left((86)/(65)\right))/(\ln 1.06)\\t & = 4.804621116\end{aligned}`

Therefore, Kevin will have paid $43,000 on the loan in 4 years and 10 months.

Question 8

Given:

• A = $3,200
• P = $1,600
• r = 2.4% = 0.024
• n = 12
• t = time (in years)

`\begin{aligned}1600\left(1+(0.024)/(12)\right)^(12t) & = 3200\\1600(1.002)^(12t) & = 3200\\(1.002)^(12t) & = (3200)/(1600)\\(1.002)^(12t) & = 2\\\ln (1.002)^(12t) & = \ln 2\\12t \ln (1.002) & = \ln 2\\12t & = (\ln 2)/(\ln 1.002)\\t & = (1)/(12)\left((\ln 2)/(\ln 1.002)\right)\\t & = 28.91000404\end{aligned}`

The balance of the account will be $3,200 in 28 years and 11 months.

Answer 2

6. 14.9 years

7. 4.8 years

8. 28.91 years

Explanation:

Exponential growth or decay is described by the function ...

y = a·b^t

where 'a' is the initial value, and 'b' is the growth (or decay) factor. The growth (or decay) factor is one added to the growth rate.

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This equation can be solved for t to get ...

y/a = b^t . . . . . . . divide by a

log(y/a) = t·log(b) . . . . take logarithms

t = log(y/a)/log(b) . . . . divide by the coefficient of t

Note that the time period is in the same units as the growth rate. (If the growth rate is per quarter, then the value of t will be in quarters.)

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6.

The initial value is a=8000; the final value is y=35000; and the growth rate is 10%/4 = 0.025 per quarter. The above equation will give the time period in quarters.

t = log(35000/8000)/log(1+0.025) ≈ 59.77 . . . quarters

Anisha will have $35,000 after 59.77/4 = 14.9 years.

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7.

No loan payment period is given, so we assume that the $43000 represents the repayment of the entire loan. The growth rate is 6% per year. The time required for the loaned amount to accumulate to a value of $43,000 is ...

t = log(43,000/32,500)/log(1 +0.06) ≈ 4.80 . . . years

If the loan period is 4.80 years, Kevin will have paid 43,000 to pay it off.

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8.

The growth rate is 0.024/12 = 0.002 per month. The number of months to achieve the desired balance is ...

t = log(3200/1600)/log(1 +0.024/12) ≈ 346.92 . . . . months

The balance in the account will be $3200 after 346.92/12 = 28.91 years.

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Additional comment

As a crude check on the answers, the "rule of 72" can be used. It says the doubling time is approximately 72 divided by the growth rate in percent. That is, a 10% growth rate will cause the value to double in approximately 72/10 = 7.2 years.

In the first problem, the value grows from 8000 to 35000, which represents a little more than 2 doublings (8000 to 16000 and 16000 to 32000). So, it is reasonable to expect the time period to be near 2 times 7.2 years, or 14.4 years. The value of 14.9 years is not unreasonable.