To solve this problem, we need to determine the continuous interest rate \(\( r \)\) required for Bilquis' investment to grow from \($560 to $\)940 over 12 years. We will use the formula for continuous compounding:
\(\[ A = P \cdot e^{rt} \]\)
where:
- \(\( A \)\) is the amount of money accumulated after n years, including interest.
- \(\( P \)\) is the principal amount (the initial amount of money).
- \(\( r \)\) is the annual interest rate (in decimal form).
- \(\( t \)\) is the time the money is invested for, in years.
- \(\( e \)\) is the base of the natural logarithm, approximately equal to 2.71828.
Given:
- \(\( A = 940 \)\)
- \(\( P = 560 \)\)
- \(\( t = 12 \)\)
We need to solve for \(\( r \)\). Rearrange the formula to isolate \(\( r \)\):
\(\[ 940 = 560 \cdot e^{12r} \]\)
First, divide both sides by 560:
\(\[ \frac{940}{560} = e^{12r} \]\)
\(\[ 1.67857 \approx e^{12r} \]\) (since 940/560 ≈ 1.67857)
Next, apply the natural logarithm (ln) to both sides to get rid of the exponential \(\( e \)\):
\(\[ \ln(1.67857) = 12r \]\)
The natural logarithm of 1.67857 is approximately:
\(\[ \ln(1.67857) \approx 0.5166 \]\)
So:
\(\[ 0.5166 = 12r \]\)
Now, solve for \(\( r \)\):
\(\[ r = \frac{0.5166}{12} \]\)
\(\[ r \approx 0.04305 \]\)
Convert \(\( r \)\) to a percentage by multiplying by 100:
\(\[ r \times 100 = 4.305 \]\)
Rounding to the nearest tenth of a percent gives:
\(\[ r \approx 4.3 \% \]\)
Therefore, the interest rate required for Bilquis to end up with $940 after 12 years with continuous compounding is 4.3% per year.
