Question 6, 3.4.19

Use the formula for the present value of an ordinary annuity or the amortization formula to solve the following problem.

Given:
[tex]\[
PV = \$19,000, \quad i = 0.005, \quad PMT = \$500
\][/tex]

Find:
[tex]\[
n = \square \quad \text{(Round up to the nearest integer.)}
\][/tex]



Answer :

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]