How long it will take for an investment of 1000 dollars to double in value if the interest rate is 9.5 percent per year, compounded continuously?
1000 e^(.095t) = 2000
e^.095t = 2
ln both sides
ln (e^(.095t)) = ln 2
.095t lne = ln2, but remember lne = 1
.095t = ln2
t = ln2/.095 = appr 7.3 years
To determine how long it will take for an investment to double in value, we can use the formula for continuous compound interest:
A = P * e^(r*t)
Where:
A is the future value of the investment
P is the principal amount (initial investment)
r is the annual interest rate
t is the time (in years)
e is the base of the natural logarithm, approximately equal to 2.71828
In this case, we know the following values:
P = $1000 (initial investment)
r = 9.5% = 0.095 (expressed as a decimal)
Our goal is to find t, the time it takes for the investment to double in value. Since we know that the future value (A) will be twice the initial investment (2P), we can rewrite the equation as:
2P = P * e^(r*t)
Dividing both sides by P, we get:
2 = e^(r*t)
Next, we need to isolate t. We can do this by taking the natural logarithm (ln) of both sides of the equation:
ln(2) = ln(e^(r*t))
Using the property of logarithms that states ln(a^b) = b*ln(a), we can simplify the equation:
ln(2) = r*t * ln(e)
Since ln(e) = 1, the equation becomes:
ln(2) = r*t
Finally, solving for t, we divide both sides by r:
t = ln(2) / r
Plugging in the values we know:
t = ln(2) / 0.095
Using a calculator, we find that ln(2) ≈ 0.693.
Substituting this value, we get:
t ≈ 0.693 / 0.095
Solving this equation, we find that t ≈ 7.3 years.
Therefore, it will take approximately 7.3 years for an investment of $1000 to double in value at an interest rate of 9.5% per year, compounded continuously.