To determine the correct expression for calculating the number of payments Jacob will need to make, we need to work with the formula for the present value of an annuity:
[tex]\[ P = \left(\frac{PMT}{i}\right) \left(1 - (1 + i)^{-n}\right) \][/tex]
Here:
- [tex]\( P \)[/tex] is the principal (loan amount), which is [tex]$950.
- \( PMT \) is the monthly payment, which is $[/tex]62.50.
- [tex]\( i \)[/tex] is the monthly interest rate, derived from the APR (Annual Percentage Rate).
- [tex]\( n \)[/tex] is the number of payments.
First, we need to convert the APR to the monthly interest rate:
[tex]\[ i = \frac{APR}{100 \times 12} = \frac{13.2}{100 \times 12} = \frac{13.2}{1200} = 0.011 \][/tex]
Now, we rearrange the original formula to solve for [tex]\( n \)[/tex] (the number of payments):
[tex]\[ P = \left(\frac{PMT}{i}\right) \left(1 - (1 + i)^{-n}\right) \][/tex]
Isolate [tex]\( (1 + i)^{-n} \)[/tex]:
[tex]\[ \frac{P}{\frac{PMT}{i}} = 1 - (1 + i)^{-n} \][/tex]
Simplify:
[tex]\[ \frac{P \cdot i}{PMT} = 1 - (1 + i)^{-n} \][/tex]
Rearrange to solve for [tex]\( (1 + i)^{-n} \)[/tex]:
[tex]\[ (1 + i)^{-n} = 1 - \frac{P \cdot i}{PMT} \][/tex]
Take the natural logarithm of both sides to solve for [tex]\( -n \)[/tex]:
[tex]\[ \log((1 + i)^{-n}) = \log\left(1 - \frac{P \cdot i}{PMT}\right) \][/tex]
[tex]\[ -n \log(1 + i) = \log\left(1 - \frac{P \cdot i}{PMT}\right) \][/tex]
Solve for [tex]\( n \)[/tex]:
[tex]\[ n = \frac{-\log\left(1 - \frac{P \cdot i}{PMT}\right)}{\log(1 + i)} \][/tex]
Since [tex]\(i\)[/tex] is positive, the expression inside the logarithm becomes:
[tex]\[ n = \frac{\log\left(\frac{PMT}{PMT - P \cdot i}\right)}{\log(1 + i)} \][/tex]
Let's match this with the given options. Using our values:
- [tex]\( P = 950 \)[/tex]
- [tex]\( PMT = 62.5 \)[/tex]
- [tex]\( i = 0.011 \)[/tex]
Matching this with the correct option:
[tex]\[ \frac{\log\left(\frac{62.5}{62.5 - 950 \times 0.011}\right)}{\log(1 + 0.011)} \][/tex]
This corresponds to option B:
[tex]\[ B. \frac{\log \left(\frac{62.5}{62.5-(950)(0.011)}\right)}{\log (1+0.011)} \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{B} \][/tex]
