You want to make an investment in a continuously-compounding account over a period of three 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.
is it 2.3
1 e^(3r) = 2
take ln of both sides
3r lne = ln2 , but lne = 1
3r = ln2
r = ln2/3 = .231..
So you will need a continuous rate of growth of 23.1 %
good luck on that one.
To determine the interest rate required for an investment to double in a given time period, we can use the formula for continuous compound interest:
A = P * e^(rt)
Where:
A = the final amount (in this case, double the initial investment)
P = the initial investment
e = the mathematical constant approximately equal to 2.71828
r = the interest rate (continuous compounding)
t = the time period
In this case, we know that we want the investment to double in three years, so t = 3. We need to solve for r.
Let's denote the final amount as 2P.
2P = P * e^(3r)
Now, divide both sides by P:
2 = e^(3r)
To isolate the exponent, we take the natural logarithm (ln) of both sides:
ln(2) = 3r
To solve for r, divide both sides by 3:
r = ln(2) / 3
Using a calculator to evaluate ln(2) and dividing by 3, we find:
r ≈ 0.231
Rounding to the nearest tenth, the interest rate required for your investment to double in three years is approximately 0.2 or 2.3%.