Question

question
a person invested $2,200 in an account growing at a rate allowing the money to double every 10 years. how long, to the nearest tenth of a year would it take for the value of the account to reach $10,800?

Explanation:

Step1: Identify the growth model

This is exponential growth with doubling time. The formula for exponential growth with doubling time \( T \) is \( A = P \times 2^{\frac{t}{T}} \), where \( A \) is the final amount, \( P \) is the principal, \( t \) is time, and \( T \) is doubling time.

Step2: Plug in values

We know \( P = 2200 \), \( A = 10800 \), \( T = 10 \). So:
\( 10800 = 2200 \times 2^{\frac{t}{10}} \)

Step3: Solve for \( \frac{t}{10} \)

Divide both sides by 2200:
\( \frac{10800}{2200} = 2^{\frac{t}{10}} \)
Simplify \( \frac{10800}{2200} = \frac{54}{11} \approx 4.909 \)
So \( 4.909 = 2^{\frac{t}{10}} \)

Step4: Take logarithm

Take log base 2 of both sides:
\( \log_2(4.909) = \frac{t}{10} \)
We know \( \log_2(x) = \frac{\ln(x)}{\ln(2)} \), so:
\( \frac{\ln(4.909)}{\ln(2)} = \frac{t}{10} \)

Step5: Calculate \( t \)

\( \ln(4.909) \approx 1.591 \), \( \ln(2) \approx 0.693 \)
\( \frac{1.591}{0.693} \approx 2.296 \)
Then \( t = 10 \times 2.296 \approx 22.96 \)
Round to nearest tenth: \( t \approx 23.0 \)

Answer:

\( 23.0 \) years