jon deposited $2,000.00 into a new savings account that earns interest compounded continuously. after 9…

jon deposited $2,000.00 into a new savings account that earns interest compounded continuously. after 9 years, the balance in the account was $7,190.00. what was the interest rate on the account? round your answer to the nearest tenth of a percent.

jon deposited $2,000.00 into a new savings account that earns interest compounded continuously. after 9 years, the balance in the account was $7,190.00. what was the interest rate on the account? round your answer to the nearest tenth of a percent.

Answer

Explanation:

Step1: Recall continuous - compounding formula

The formula for continuous - compounding is $A = Pe^{rt}$, where $A$ is the final amount, $P$ is the principal amount, $r$ is the annual interest rate (in decimal), and $t$ is the time in years. We know that $P=$2000$, $A = $7190$, and $t = 9$ years. Substitute these values into the formula: $7190=2000e^{9r}$.

Step2: Isolate the exponential term

Divide both sides of the equation by 2000: $\frac{7190}{2000}=e^{9r}$, so $3.595 = e^{9r}$.

Step3: Take the natural logarithm of both sides

Since $\ln(e^{x})=x$, taking the natural logarithm of both sides gives $\ln(3.595)=\ln(e^{9r})$. So, $\ln(3.595) = 9r$. We know that $\ln(3.595)\approx1.279$. Then $1.279 = 9r$.

Step4: Solve for $r$

Divide both sides by 9: $r=\frac{1.279}{9}\approx0.142$.

Step5: Convert $r$ to a percentage

To convert $r$ from a decimal to a percentage, multiply by 100: $r = 0.142\times100 = 14.2%$.

Answer:

$14.2$