TabaSue invested \( \$ 11,550 \) dollars in a savings account that paid \( 5.3 \% \) interest compounded annually. Write the exponential equation \( A \) that represents TabaSue's investment where \( A \) is the accrued value of her savings and \( t \) is the time of the investment in years. \( A= \) Determine how much money TabaSue will have after 11 years. Answer: After 11 years, TabaSue will have the nearest penny/cent.) Determine how long it will take for TabaSue's investment to double. Answer: TabaSue's investment will have doubled in value after in her savings account. (Round to nearest tenth.)

To represent TabaSue's investment with an exponential equation, we can use the formula for compound interest: \[ A = P(1 + r)^t \] where: - \( A \) is the accrued value of the investment, - \( P \) is the principal amount (initial investment), - \( r \) is the annual interest rate (as a decimal), - \( t \) is the time in years. Given: - \( P = 11,550 \) - \( r = 5.3\% = 0.053 \) The exponential equation representing TabaSue's investment is: \[ A = 11550(1 + 0.053)^t \] This simplifies to: \[ A = 11550(1.053)^t \] ### Determine how much money TabaSue will have after 11 years. Now, we can calculate \( A \) after \( t = 11 \) years: \[ A = 11550(1.053)^{11} \] Calculating \( (1.053)^{11} \): \[ (1.053)^{11} \approx 1.7137 \] Now, substituting this value back into the equation: \[ A \approx 11550 \times 1.7137 \approx 19876.44 \] So, after 11 years, TabaSue will have approximately: \[ \text{Answer: } \$19,876.44 \] ### Determine how long it will take for TabaSue's investment to double. To find out how long it will take for the investment to double, we set \( A = 2P \): \[ 2P = 11550(1.053)^t \] This means: \[ 23100 = 11550(1.053)^t \] Dividing both sides by 11550: \[ 2 = (1.053)^t \] Now, we take the logarithm of both sides: \[ \log(2) = t \cdot \log(1.053) \] Solving for \( t \): \[ t = \frac{\log(2)}{\log(1.053)} \] Calculating the logarithms: \[ \log(2) \approx 0.3010 \] \[ \log(1.053) \approx 0.0225 \] Now substituting these values: \[ t \approx \frac{0.3010}{0.0225} \approx 13.4 \] So, TabaSue's investment will have doubled in value after approximately: \[ \text{Answer: } 13.4 \text{ years} \]