Suppose that $2500 is invested at an interest rate of 2.5% per year, compounded continuously. After how many years will the initial investment be doubled?
To find out how many years it will take for the initial investment to double, we can use the continuous compound interest formula:
A = Pe^rt
Where:
A = Final amount or future value
P = Principal amount or initial investment
e = Euler's number (approximately 2.71828)
r = Interest rate per year
t = Time in years
Since we want to find out when the initial investment doubles, the final amount will be 2P (twice the initial investment). Let's substitute the given values into the formula:
2P = Pe^(0.025t)
Now, simplify the equation by canceling out the P:
2 = e^(0.025t)
Next, take the natural logarithm (ln) of both sides to isolate the exponent:
ln(2) = ln(e^(0.025t))
The natural logarithm of e simplifies to 1:
ln(2) = 0.025t
Now, divide both sides by 0.025 to solve for t:
t = ln(2) / 0.025
Calculating this, we find:
t ≈ 27.726 years
Therefore, it will take approximately 27.726 years for the initial investment to double.
To determine the number of years it takes for an investment to double at a given interest rate compounded continuously, we can use the formula:
\[A = P \cdot e^{rt}\]
Where:
- A is the final amount (double the initial investment)
- P is the initial investment ($2500)
- e is a mathematical constant approximately equal to 2.71828
- r is the interest rate per year (2.5% or 0.025 in decimal form)
- t is the number of years we want to find
Since we want to find the number of years for the initial investment to double, we'll set A equal to 2 times the initial investment (2P):
\[2P = P \cdot e^{rt}\]
Next, we can simplify the equation by dividing both sides by P:
\[2 = e^{rt}\]
Now, we can take the natural logarithm (ln) of both sides to isolate the exponent:
\[ln(2) = rt \cdot ln(e)\]
Simplifying further, we know that ln(e) is equal to 1:
\[ln(2) = rt\]
We can now solve for t by dividing both sides by r and rearranging the equation:
\[t = \frac{ln(2)}{r}\]
Substituting the values, we have:
\[t = \frac{ln(2)}{0.025}\]
Calculating this expression, we find:
\[t \approx \frac{0.693}{0.025} \approx 27.72 \text{ years}\]
Therefore, the initial investment will be doubled after approximately 27.72 years.
The formula to use can be found and is derived here:
http://www.moneychimp.com/articles/finworks/continuous_compounding.htm
In your case, the answer is obtained by solving
2500*e^0.025Y = 5000
e^.025Y = 2
Y is the number of years and e = 2.71828...
0.025Y = ln2 = 0.69315
Y = 27.7 years