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Sagot :
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.
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]
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