Math  /  Algebra

Question-• www-awu.aleks.com/alekscgi/x/Isl.exe/10_u-lgNslkr7j8P3jH-lis1WHQv7rHh0j2icTeSIC4QOKrKw0C-WYA3ZDkXtcq7DqYNM1OKbSqSQ4wahNgLZp_78rEH30JXTdqk... Exponential and Logarithmic Functions Finding the time in a word problem on compound interest Morg:
Teresa needs $6260\$ 6260 for a future project. She can invest $4000\$ 4000 now at an annual rate of 10.8%10.8 \%, compounded monthly. Assuming that no withdrawals are made, how long will it take for her to have enough money for her project?
Do not round any intermediate computations, and round your answer to the nearest hundredth. \square years

Studdy Solution

STEP 1

1. Teresa needs a future amount of 6260.<br/>2.Shecaninvest6260.<br />2. She can invest 4000 now.
3. The annual interest rate is 10.8%, compounded monthly.
4. No withdrawals are made during the investment period.
5. We need to find the time it will take for the investment to grow to $6260.

STEP 2

1. Identify the formula for compound interest.
2. Define the variables in the formula.
3. Substitute the known values into the formula.
4. Solve for the unknown variable, which is time.

STEP 3

Identify the formula for compound interest. The formula is:
A=P(1+rn)nt A = P \left(1 + \frac{r}{n}\right)^{nt}
where: - A A is the future value of the investment/loan, including interest. - P P is the principal investment amount (initial deposit or loan amount). - r r is the annual interest rate (decimal). - n n is the number of times that interest is compounded per year. - t t is the number of years the money is invested for.

STEP 4

Define the variables in the formula based on the problem:
- A=6260 A = 6260 - P=4000 P = 4000 - r=10.8%=0.108 r = 10.8\% = 0.108 - n=12 n = 12 (since the interest is compounded monthly)

STEP 5

Substitute the known values into the formula:
6260=4000(1+0.10812)12t 6260 = 4000 \left(1 + \frac{0.108}{12}\right)^{12t}

STEP 6

Solve for the unknown variable, t t .
First, simplify the expression inside the parentheses:
1+0.10812=1+0.009=1.009 1 + \frac{0.108}{12} = 1 + 0.009 = 1.009
Now, substitute back into the equation:
6260=4000×(1.009)12t 6260 = 4000 \times (1.009)^{12t}
Divide both sides by 4000 to isolate the exponential term:
62604000=(1.009)12t \frac{6260}{4000} = (1.009)^{12t}
Calculate the left side:
1.565=(1.009)12t 1.565 = (1.009)^{12t}
Take the natural logarithm of both sides to solve for t t :
ln(1.565)=ln((1.009)12t) \ln(1.565) = \ln((1.009)^{12t})
Using the logarithmic identity ln(ab)=bln(a) \ln(a^b) = b \cdot \ln(a) , we have:
ln(1.565)=12tln(1.009) \ln(1.565) = 12t \cdot \ln(1.009)
Solve for t t :
t=ln(1.565)12ln(1.009) t = \frac{\ln(1.565)}{12 \cdot \ln(1.009)}
Calculate the value of t t :
t0.44715812×0.008961 t \approx \frac{0.447158}{12 \times 0.008961}
t0.4471580.107532 t \approx \frac{0.447158}{0.107532}
t4.16 t \approx 4.16
The time it will take for Teresa's investment to grow to $6260 is approximately \( \boxed{4.16} \) years.

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