You want to make an investment in a continuously compounding account earning 12.6% interest. How many years will it take for your investment to double in value? Round the natural log value to the nearest thousandth. Round the answer to the nearest year.
I got 5 years
2 = e^(.126 t)
ln(2) = .126 t
5.501... , should round up
P = Po*e^rt = 2Po.
Po*e^rt = 2Po,
e^rt = 2,
rt*Lne = Ln2,
rt = Ln2/Lne,
r*t = 0.69,
t = 0.69/0.126 = 5.5 yrs.
To find out how many years it will take for your investment to double in value, you can use the formula for continuous compounding:
A = P * e^(r * t)
Where:
A is the final amount (double the initial investment value)
P is the initial investment value
r is the interest rate
t is the time in years
e is the base of the natural logarithm
In this case, you want to double your investment, so A = 2 * P. The interest rate is given as 12.6%, which can be written as 0.126 in decimal form. Rearranging the formula, we obtain:
2P = P * e^(0.126 * t)
Dividing both sides of the equation by P, we get:
2 = e^(0.126 * t)
To solve for t, we can take the natural log (ln) of both sides of the equation, since the ln function is the inverse of the exponential function:
ln(2) = ln(e^(0.126 * t))
By the properties of logarithms, this simplifies to:
ln(2) = 0.126 * t
Now we can solve for t by dividing both sides of the equation by 0.126:
t = ln(2) / 0.126
Using a calculator to evaluate this expression, we get:
t ≈ 5.518
So it will take approximately 5.518 years for your investment to double in value. Rounded to the nearest year, this would be 6 years. Note that the natural log value should be rounded to the nearest thousandth, which would be 0.693.