To solve this problem, let's use the formula for compound interest:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Where:
- [tex]\( A \)[/tex] is the final amount of money.
- [tex]\( P \)[/tex] is the principal amount (initial investment).
- [tex]\( r \)[/tex] is the annual interest rate (as a decimal).
- [tex]\( n \)[/tex] is the number of times interest is compounded per year.
- [tex]\( t \)[/tex] is the time the money is invested for, in years.
Given:
- Principal amount, [tex]\( P \)[/tex].
- Annual interest rate, [tex]\( r = 12\% = 0.12 \)[/tex].
- The money is compounded monthly, so [tex]\( n = 12 \)[/tex].
- We want to find the time [tex]\( t \)[/tex] it will take for the principal to become four times its original value, so [tex]\( A = 4P \)[/tex].
First, let's rewrite the formula with these values:
[tex]\[ 4P = P \left(1 + \frac{0.12}{12}\right)^{12t} \][/tex]
Next, we can simplify the equation by dividing both sides by [tex]\( P \)[/tex]:
[tex]\[ 4 = \left(1 + \frac{0.12}{12}\right)^{12t} \][/tex]
Now, calculate [tex]\( \frac{0.12}{12} \)[/tex]:
[tex]\[ \frac{0.12}{12} = 0.01 \][/tex]
So the equation becomes:
[tex]\[ 4 = (1 + 0.01)^{12t} \][/tex]
[tex]\[ 4 = (1.01)^{12t} \][/tex]
To solve for [tex]\( t \)[/tex], let's take the natural logarithm of both sides:
[tex]\[ \ln(4) = \ln((1.01)^{12t}) \][/tex]
Using the logarithm property [tex]\( \ln(a^b) = b \ln(a) \)[/tex]:
[tex]\[ \ln(4) = 12t \cdot \ln(1.01) \][/tex]
Now, solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(4)}{12 \cdot \ln(1.01)} \][/tex]
After calculating the natural logarithms:
[tex]\[ \ln(4) \approx 1.386 \][/tex]
[tex]\[ \ln(1.01) \approx 0.009950 \][/tex]
So:
[tex]\[ t = \frac{1.386}{12 \cdot 0.009950} \][/tex]
[tex]\[ t \approx \frac{1.386}{0.1194} \][/tex]
[tex]\[ t \approx 11.61 \][/tex]
Therefore, it will take approximately 11.61 years for the principal to become four times its original value.
The correct answer is:
b. 11.61 years
