With interest at 9% compounded annually, what is the fewest number of years
The correct answer and explanation is:
To determine the fewest number of years required for an investment to grow at 9% annual compound interest, you will need to apply the compound interest formula: A=P(1+rn)ntA = P \left( 1 + \frac{r}{n} \right)^{nt}
Where:
- AA is the amount of money accumulated after interest.
- PP is the principal amount (the initial investment).
- rr is the annual interest rate (in decimal form).
- nn is the number of times the interest is compounded per year.
- tt is the number of years.
In this case, the interest is compounded annually, meaning n=1n = 1, and the rate is 9%, or r=0.09r = 0.09.
If the goal is to double the investment, the equation becomes: 2P=P(1+0.09)t2P = P \left( 1 + 0.09 \right)^{t}
Simplifying: 2=(1.09)t2 = (1.09)^{t}
Now, solve for tt by taking the natural logarithm of both sides: ln(2)=tln(1.09)\ln(2) = t \ln(1.09) t=ln(2)ln(1.09)t = \frac{\ln(2)}{\ln(1.09)}
Using logarithmic values: t≈0.69310.0862≈8.03t \approx \frac{0.6931}{0.0862} \approx 8.03
Since the number of years must be a whole number, the fewest number of years required for an investment to double at a 9% interest rate, compounded annually, is 9 years.
This result is based on the assumption that the investment is compounded annually. The process demonstrates how exponential growth leads to a doubling of the investment, with the time for such a growth depending on the interest rate and the frequency of compounding. In practice, the compound interest allows the investment to grow progressively faster as time goes on.