Question

an investment triples every 12 years.
what is the annual interest rate (compounded continuously)?
round to the ten - thousandths place.
r=

question 4
1 pts
a bacteria population doubles every 3 hours.
if 500 bacteria are present initially, how many after 12 hours?

Explanation:

Step1: Recall the continuous - compounding formula

The formula for continuous compounding is \(A = P e^{rt}\), where \(A\) is the final amount, \(P\) is the principal amount, \(r\) is the annual interest rate, and \(t\) is the time in years.

Step2: Substitute the given values into the formula

Given that the investment triples, so \(A = 3P\) and \(t = 12\). Substituting into \(A=Pe^{rt}\), we get \(3P=Pe^{12r}\).

Step3: Simplify the equation

Divide both sides of the equation \(3P = Pe^{12r}\) by \(P\) (since \(P
eq0\)). We obtain \(3=e^{12r}\).

Step4: Take the natural logarithm of both sides

Using the property \(\ln(e^{x})=x\), if \(3 = e^{12r}\), then \(\ln(3)=\ln(e^{12r})\). So \(\ln(3)=12r\).

Step5: Solve for \(r\)

We know that \(\ln(3)\approx1.0986\). Then \(r=\frac{\ln(3)}{12}\). Substituting \(\ln(3)\approx1.0986\) into the formula, \(r=\frac{1.0986}{12}\).

Step6: Calculate the value of \(r\)

\(r=\frac{1.0986}{12}=0.0916\)

Answer:

\(r = 0.0916\)