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Math  /  Algebra

Question19. If PP dollars are invested at the end of each year in an annuity that earns interest at an annual rate rr, the amount in the account will be AA dollars after nn years, where n=log⁡[ArP+1]log⁡(1+r)n=\frac{\log \left[\frac{A r}{P}+1\right]}{\log (1+r)}
If $5,000\$ 5,000 is invested each year in an annuity earning 8%8 \% annual interest, when will the account be worth $75,000\$ 75,000 ? Round to the nearest tenth.

Studdy Solution

STEP 1

1. The annual investment amount P P is \5,000.
2. The annual interest rate \( r \) is 8%, or 0.08 in decimal form.
3. The desired account balance \( A \) is \$75,000.
4. We need to find the number of years \( n \) it will take for the account to reach \$75,000.
5. We will use the given formula for \( n \): n = \frac{\log \left[\frac{A r}{P} + 1\right]}{\log (1+r)} \]

STEP 2

1. Substitute the given values into the formula.
2. Calculate the intermediate value inside the logarithm.
3. Compute the logarithms.
4. Solve for n n and round to the nearest tenth.

STEP 3

Substitute the given values into the formula: n=log⁡[75,000×0.085,000+1]log⁡(1+0.08)n = \frac{\log \left[\frac{75,000 \times 0.08}{5,000} + 1\right]}{\log (1 + 0.08)}

STEP 4

Calculate the intermediate value inside the logarithm: 75,000×0.085,000=6,0005,000=1.2\frac{75,000 \times 0.08}{5,000} = \frac{6,000}{5,000} = 1.2 So the expression becomes: n=log⁡(1.2+1)log⁡(1.08)n = \frac{\log(1.2 + 1)}{\log(1.08)}

STEP 5

Compute the logarithms: log⁡(2.2)andlog⁡(1.08)\log(2.2) \quad \text{and} \quad \log(1.08)
Using a calculator: log⁡(2.2)≈0.3424\log(2.2) \approx 0.3424 log⁡(1.08)≈0.0334\log(1.08) \approx 0.0334

STEP 6

Solve for n n : n=0.34240.0334≈10.25n = \frac{0.3424}{0.0334} \approx 10.25
Round to the nearest tenth: n≈10.3n \approx 10.3
The account will be worth \$75,000 in approximately \( \boxed{10.3} \) years.

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