Respuesta :
Answer:
James will pay off his mortgage 136 months early.
Step-by-step explanation:
Given : Amount to be paid(A)= $205000
Interest rate per annum = 5.78% = 0.0578
monthly interest rate(i) = [tex]\frac{0.0578}{12}[/tex]= 0.004817 (approx)
Monthly payment (P) = $1500.
we have to find how many months early will he pay off his mortgage
We know, [tex]P=\frac{A(i)}{1-(1+i)^(-t)}[/tex]
Substitute values , we get,
[tex]1500=\frac{205000(0.004817)}{1-(1+0.004817)^(-t})[/tex]
[tex]1500=\frac{205000(0.004817)}{1-(1.004817)^(-t)}[/tex]
[tex]1500=\frac{987.416653}{(1-(1.004817)^(-t)}[/tex]
Solving for t ,
[tex]{(1-(1.004817)^t)}=\frac{987.416653}{1500}\\\\\\{(1-(1.004817)^(-t))}=0.65828\\\\\(1.004817)^(-t)=0.341722[/tex]
Applying ln both sides,we get,
[tex]\ln \left(1.004817^(-t)\right)=\ln \left(0.341722\right)[/tex]
[tex]\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)[/tex]
[tex]-t\ln \left(1.004817\right)=\ln \left(0.341722\right)[/tex]
[tex]\mathrm{Divide\:both\:sides\:by\:}\ln \left(1.004817\right)[/tex]
[tex]\frac{-t\ln \left(1.004817\right)}{\ln \left(1.004817\right)}=\frac{\ln \left(0.341722\right)}{\ln \left(1.004817\right)}[/tex]
[tex]-t=\frac{\ln \left(0.341722\right)}{\ln \left(1.004817\right)}[/tex]
[tex]t=223.44650\dots[/tex]
Thus, t = 224 months.
1 year = 12 month thus, 30 years = 360 months.
Difference = 360 - 224 = 136 months
Thus, James will pay off his mortgage (360-224) 136 months early.
