Question

if $1000 is deposited in an account paying 7% annual interest, compounded continuously, how long will it take for the account to increase to $1800? round your answer to the nearest hundredth of a year. about years

Explanation:

Step1: Write the 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 form), and \(t\) is the time in years.

Given \(P=\$1000\), \(A = \$1800\), and \(r=0.07\). Substitute these values into the formula: \(1800=1000e^{0.07t}\).

Step2: Simplify the equation

Divide both sides of the equation by \(1000\): \(\frac{1800}{1000}=e^{0.07t}\), so \(1.8 = e^{0.07t}\).

Step3: Take the natural logarithm of both sides

Using the property \(\ln(e^{x})=x\), if \(y = e^{x}\), then \(x=\ln(y)\).

Take the natural logarithm of both sides: \(\ln(1.8)=\ln(e^{0.07t})\). Since \(\ln(e^{0.07t}) = 0.07t\), the equation becomes \(\ln(1.8)=0.07t\).

Step4: Solve for \(t\)

We know that \(\ln(1.8)\approx0.587787\). Then \(t=\frac{\ln(1.8)}{0.07}\).

Substitute \(\ln(1.8)\approx0.587787\) into the formula: \(t=\frac{0.587787}{0.07}\approx8.40\).

Answer:

\(8.40\)