To solve for the number of periods [tex]\( n \)[/tex] in an amortization problem, we use the amortization formula:
[tex]\[
PV = PMT \times \left( \frac{1 - (1 + i)^{-n}}{i} \right)
\][/tex]
Given:
- Present Value (PV) = [tex]$19,000
- Interest rate per period (i) = 0.005
- Payment per period (PMT) = $[/tex]500
We need to solve for the number of periods [tex]\( n \)[/tex]. To do this, we isolate [tex]\( n \)[/tex] in the formula.
First, let's rewrite the formula to solve for [tex]\( (1 + i)^{-n} \)[/tex]:
[tex]\[
\frac{PV \times i}{PMT} = 1 - (1 + i)^{-n}
\][/tex]
Substitute the given values:
[tex]\[
\frac{19,000 \times 0.005}{500} = 1 - (1 + 0.005)^{-n}
\][/tex]
Calculate the left-hand side:
[tex]\[
\frac{95}{500} = 1 - (1.005)^{-n}
\][/tex]
[tex]\[
0.19 = 1 - (1.005)^{-n}
\][/tex]
Rearrange to isolate the term with [tex]\( n \)[/tex]:
[tex]\[
(1.005)^{-n} = 1 - 0.19
\][/tex]
[tex]\[
(1.005)^{-n} = 0.81
\][/tex]
Now, to solve for [tex]\( -n \)[/tex], we take the natural logarithm (log) of both sides:
[tex]\[
\ln((1.005)^{-n}) = \ln(0.81)
\][/tex]
Using the properties of logarithms [tex]\( \ln(a^b) = b \ln(a) \)[/tex]:
[tex]\[
-n \ln(1.005) = \ln(0.81)
\][/tex]
Solve for [tex]\( n \)[/tex]:
[tex]\[
n = \frac{\ln(0.81)}{-\ln(1.005)}
\][/tex]
Calculating the logarithms and the division:
[tex]\[
\ln(0.81) \approx -0.21072103131565256
\][/tex]
[tex]\[
\ln(1.005) \approx 0.004987541511038968
\][/tex]
[tex]\[
n = \frac{-0.21072103131565256}{-0.004987541511038968} \approx 42.249479197167965
\][/tex]
Since the number of periods must be an integer and we round up to the nearest integer, we get:
[tex]\[
n \approx 43
\][/tex]
So, the number of periods [tex]\( n \)[/tex] is approximately:
[tex]\[
n = 43
\][/tex]
