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