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question 8
find the time it takes for $8,500 to double when invested at an annual interest rate of 7%, compounded continuously.
years
find the time it takes for $850,000 to double when invested at an annual interest rate of 7%, compounded continuously.
years
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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 amount, $r$ is the annual interest rate (in decimal), and $t$ is the time in years.
We want to find the time it takes for the principal to double, so $A = 2P$. Let's substitute $A = 2P$ into the formula:
$$2P = Pe^{rt}$$
Step2: Simplify the equation
Divide both sides of the equation by $P$ (since $P
eq0$):
$$2 = e^{rt}$$
We know that the annual interest rate $r = 7\% = 0.07$. Substitute $r = 0.07$ into the equation:
$$2 = e^{0.07t}$$
Step3: Solve for $t$
Take the natural logarithm of both sides to solve for $t$:
$$\ln(2) = \ln(e^{0.07t})$$
Using the property of logarithms $\ln(e^x)=x$, we get:
$$\ln(2) = 0.07t$$
Now, solve for $t$ by dividing both sides by $0.07$:
$$t = \frac{\ln(2)}{0.07}$$
Step4: Calculate the value of $t$
We know that $\ln(2)\approx0.6931$. So:
$$t = \frac{0.6931}{0.07}\approx9.90$$
Notice that the principal amount ($8,500 or $850,000) does not affect the time it takes to double because in the continuous compounding formula, when we set $A = 2P$, the $P$ cancels out. So both amounts will take the same time to double.
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For $8,500: $\boxed{9.90}$ (or approximately 10) years.
For $850,000: $\boxed{9.90}$ (or approximately 10) years.