You want to make an investment in a continuously compounding account over a period of two years. What interest rate is required for your investment to double in that time period? Round the logarithm value and the answer to the nearest tenth.
suppose we invest $1.00
2 = 1 e^(2r)
ln 2 = 2r lne , but lne = 1
2r = ln2
r = ln2/2 = appr .34657..
= appr 34.66%
nice if you can get it
To find the interest rate required for your investment to double in a continuously compounding account, we can use the formula for compound interest:
A = P * e^(rt)
Where:
A = Final amount (double the initial investment)
P = Principal amount (initial investment)
e = Euler's number (approximately 2.71828)
r = Interest rate (unknown)
t = Time period in years
Since we want the final amount (A) to be double the initial investment (P), we can rewrite the formula as:
2P = P * e^(rt)
Next, we can simplify the equation by dividing both sides by P:
2 = e^(rt)
Now, take the natural logarithm (ln) of both sides of the equation:
ln(2) = ln(e^(rt))
Using the logarithmic property that ln(e^x) = x, the equation becomes:
ln(2) = rt
To find the interest rate (r), divide both sides of the equation by t:
r = ln(2) / t
Now, plug in the values. Since the time period is two years:
r = ln(2) / 2
Using a calculator, calculate the natural logarithm of 2:
ln(2) ≈ 0.69315
Now, substitute the values into the equation:
r ≈ 0.69315 / 2
Round the logarithm value and the answer to the nearest tenth:
r ≈ 0.3
Therefore, to double your investment in a continuously compounding account over a period of two years, you would need an interest rate of approximately 0.3 or 30%.