Answered

Consider the monthly payment formula M = (Pr(1+r)^n/(1+r)^n-1 What is the name of the formula you get if you solve for P?

What about if you solve for n? What special function do you need to use when solving for n?

I will give brainliest if correct.



Answer :

mhanifa

In the given formula we have variables:

  • M - monthly payment,
  • P - principal,
  • r - interest rate,
  • n - number of payments.

If we have to solve it for P then it should be called 'Principal' or 'Loan amount'.

If we have to solve it for n then it should be called 'Number of payments' or 'Number of installments'.

Since n is the power of a number we need logarithm to solve it for n.

semsee45

Answer:

"Principal (loan amount)"

"Term of the loan (in months)" or "Number of monthly payments"

Logarithms

Step-by-step explanation:

Monthly Payment Formula

[tex]M=\dfrac{Pr\left(1+r\right)^n}{\left(1+r\right)^n-1}[/tex]

where:

  • M = monthly payment.
  • P = principal loan amount.
  • r = interest rate per month (in decimal form).
  • n = term of the loan (in months).

If we solve for P, the name of the formula is "Principal (loan amount)".

If we solve for n, the name of the formula is "Term of the loan (in months)" or "Number of monthly payments".

When solving for n, we need to use logarithms:

[tex]\boxed{\begin{aligned}M&=\frac{Pr\left(1+r\right)^n}{\left(1+r\right)^n-1}\\\\M(\left(1+r\right)^n-1)&=Pr\left(1+r\right)^n\\\\\frac{\left(1+r\right)^n-1}{\left(1+r\right)^n}&=\frac{Pr}{M}\\\\1-\frac{1}{\left(r+1\right)^n}&=\frac{Pr}{M}\\\\\frac{1}{\left(r+1\right)^n}&=1-\frac{Pr}{M}\\\\\left(r+1\right)^{-n}&=1-\frac{Pr}{M}\\\\\ln \left(r+1\right)^{-n}&=\ln \left(1-\frac{Pr}{M}\right)\\\\-n\ln(r+1)&=\ln\left(1-\frac{Pr}{M}\right)\\\\n&=\frac{-\ln\left(1-\frac{Pr}{M}\right)}{\ln(r+1)}\end{aligned}}[/tex]

Other Questions