Question

1. 0 / 6.66 points

suppose you invest $6,200 in an account that pays 4.5% interest per year, compounded continuously.
(a) what is the amount (in $) after 2 years? (round your answer to the nearest cent.)
$ ×
(b) how long (in yr) will it take for the amount to be $9,000? (round your answer to two decimal places.)
× yr

Explanation:

Step1: Recall the continuous - compounding formula

The formula for continuous compounding is \(A = Pe^{rt}\), where \(P\) is the principal amount, \(r\) is the annual interest rate (in decimal form), \(t\) is the time in years, and \(A\) is the amount of money in the account after \(t\) years.

Step2: Solve part (a)

Given \(P=\$6200\), \(r = 0.045\), and \(t = 2\).
Substitute these values into the formula \(A=Pe^{rt}\):

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Using a calculator, \(e^{0.09}\approx1.094174\).
Then \(A = 6200\times1.094174=\$6783.88\)

Step3: Solve part (b)

Given \(A = 9000\), \(P = 6200\), \(r=0.045\). Substitute into \(A = Pe^{rt}\):

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Take the natural logarithm of both sides: \(\ln(\frac{45}{31})=\ln(e^{0.045t})\)
Since \(\ln(e^{x})=x\), we have \(\ln(\frac{45}{31}) = 0.045t\)
\(\ln(\frac{45}{31})\approx\ln(1.451613)\approx0.3727\)
Then \(t=\frac{\ln(\frac{45}{31})}{0.045}=\frac{0.3727}{0.045}\approx8.28\)

Answer:

(a) \(\$6783.88\)
(b) \(8.28\) yr