How long will it take for an investment of $13,000 to double if the investment earns interest at the rate of 6%/year compounded continuously?
13000 e^.06t = 26000
e^.06t = 2
take ln of both sides
ln e^.06t = ln 2
.06t ln e = ln2 , but lne = 1
t = ln2/.06
= appr 11.55
To find out how long it will take for an investment to double with continuous compounding, you can use the formula for continuous compound interest:
A = P*e^(rt)
where:
A = the final amount (double the initial investment)
P = the principal amount (initial investment)
e = Euler's number (approximately 2.71828)
r = the interest rate (in decimal form)
t = the time (in years)
In this case, P = $13,000 and r = 6%/year = 0.06/year. We need to solve for t.
Substituting these values into the formula, we get:
2P = P*e^(0.06t)
Next, we can simplify the equation by canceling out P on both sides:
2 = e^(0.06t)
To solve for t, we need to isolate the variable. Taking the natural logarithm (ln) of both sides:
ln(2) = ln(e^(0.06t))
ln(2) = 0.06t*ln(e)
Since ln(e) is equal to 1, we have:
ln(2) = 0.06t
Now, divide both sides by 0.06:
t = ln(2)/0.06
Using a calculator, we find:
t ≈ 11.55 years
Therefore, it will take approximately 11.55 years for an investment of $13,000 to double with an interest rate of 6% per year compounded continuously.