Math  /  Algebra

QuestionIf $4,500.00\$ 4,500.00 is placed into an account earning 2.8%2.8 \% interest compounded amually. how long will it take for the amount in the account to double? Round to one decimal.

Studdy Solution

STEP 1

1. The initial principal amount is \$4,500.00.
2. The interest rate is 2.8% compounded annually.
3. We want to find the time it takes for the principal to double.
4. The formula for compound interest is used.

STEP 2

1. Understand the compound interest formula.
2. Set up the equation for doubling the principal.
3. Solve for the time variable.
4. Round the result to one decimal place.

STEP 3

Understand the compound interest formula. The formula for compound interest is:
A=P(1+rn)nt A = P \left(1 + \frac{r}{n}\right)^{nt}
where: - A A is the amount of money accumulated after n years, including interest. - P P is the principal amount (initial investment). - 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 time in years.
For this problem, since the interest is compounded annually, n=1 n = 1 .

STEP 4

Set up the equation for doubling the principal. We want the final amount A A to be double the initial principal P P .
2P=P(1+r1)1t 2P = P \left(1 + \frac{r}{1}\right)^{1 \cdot t}
Since P=4500 P = 4500 and r=0.028 r = 0.028 , the equation becomes:
2×4500=4500(1+0.028)t 2 \times 4500 = 4500 \left(1 + 0.028\right)^t

STEP 5

Solve for the time variable t t .
First, divide both sides by 4500:
2=(1.028)t 2 = (1.028)^t
To solve for t t , take the natural logarithm of both sides:
ln(2)=ln((1.028)t) \ln(2) = \ln((1.028)^t)
Using the logarithmic identity ln(ab)=bln(a) \ln(a^b) = b \cdot \ln(a) , we have:
ln(2)=tln(1.028) \ln(2) = t \cdot \ln(1.028)
Now, solve for t t :
t=ln(2)ln(1.028) t = \frac{\ln(2)}{\ln(1.028)}

STEP 6

Calculate the value of t t and round to one decimal place.
Using a calculator:
t0.6931470.02763125.0 t \approx \frac{0.693147}{0.027631} \approx 25.0
Therefore, it will take approximately 25.0 \boxed{25.0} years for the amount to double.

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