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sqdancefan

Answer:

  6. 14.9 years

  7. 4.8 years

  8. 28.91 years

Step-by-step 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.

semsee45

Answer:

Compound interest formula

[tex]\Large \text{$ \sf A=P(1+\frac{r}{n})^{nt} $}[/tex]

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)

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

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)

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

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)

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

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

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