Question

question
2300 dollars is placed in an account with an annual interest rate of 5.5%. to the nearest tenth of a year, how long will it take for the account value to reach 4300 dollars?

Explanation:

Step1: Identify the formula for compound interest (assuming compound interest, as simple interest is less common for such problems without specification, and compound interest formula is \( A = P(1 + r)^t \), where \( A \) is the final amount, \( P \) is the principal, \( r \) is the annual interest rate (decimal), and \( t \) is time in years.

Given \( P = 2300 \), \( A = 4300 \), \( r = 5.5\% = 0.055 \). So we have the equation \( 4300 = 2300(1 + 0.055)^t \).

Step2: Divide both sides by 2300 to isolate the exponential term.

\( \frac{4300}{2300} = (1.055)^t \)
Simplify \( \frac{4300}{2300} \approx 1.8696 \), so \( 1.8696 = (1.055)^t \).

Step3: Take the natural logarithm of both sides.

\( \ln(1.8696) = \ln((1.055)^t) \)
Using the logarithm power rule \( \ln(a^b) = b\ln(a) \), we get \( \ln(1.8696) = t\ln(1.055) \).

Step4: Solve for \( t \) by dividing both sides by \( \ln(1.055) \).

\( t = \frac{\ln(1.8696)}{\ln(1.055)} \)
Calculate \( \ln(1.8696) \approx 0.626 \) and \( \ln(1.055) \approx 0.0535 \). Then \( t = \frac{0.626}{0.0535} \approx 11.7 \).

Answer:

\( 11.7 \) years