Question

how many years will it take for an initial investment of $10,000 to grow to $35,000? assume a rate of interest of 20% compounded continuously.
it will take about \\(\square\\) years for the investment to grow to $35,000.
(round to two decimal places as needed.)

Explanation:

Step1: Recall the continuous compounding formula

The formula for continuous compounding is $A = Pe^{rt}$, where $A$ is the final amount, $P$ is the principal (initial investment), $r$ is the annual interest rate (in decimal), and $t$ is the time in years.
We know $P = 10000$, $A = 35000$, and $r = 0.20$ (since 20% = 0.20). We need to solve for $t$.
Substitute the known values into the formula: $35000 = 10000e^{0.20t}$

Step2: Simplify the equation

Divide both sides of the equation by 10000: $\frac{35000}{10000}=e^{0.20t}$
Simplify the left side: $3.5 = e^{0.20t}$

Step3: Take the natural logarithm of both sides

Take the natural log (ln) of both sides: $\ln(3.5)=\ln(e^{0.20t})$
Using the property of logarithms $\ln(e^x)=x$, we get: $\ln(3.5) = 0.20t$

Step4: Solve for t

Divide both sides by 0.20: $t=\frac{\ln(3.5)}{0.20}$
Calculate $\ln(3.5)\approx1.252763$, then $t=\frac{1.252763}{0.20}\approx6.263815$

Answer:

6.26