Depreciation:

A construction company purchases a bulldozer for $160,000. Each year the value of the buldozer depreciates by 20% of its value in the preceeding year. Let V sub n be the value of the bulldozer.(Let n=1 be the year the bulldozer purchase.)
a.) Find the Formula.
b.) In what year will the value of the bulldozer be less than $100,000?

each year its worth is 80% = 0.8 of its previous value. If Vn is the value at the end of year n, then

Vn = 160000 * 0.8^n

4 years

a.) To find the formula for the value of the bulldozer, we can use the concept of compound interest, where the value depreciates by 20% each year.

Let Vn be the value of the bulldozer in year n.
V1 = $160,000 (initial value)

In year 2, the value will be 80% of the previous year's value: V2 = 0.8 * V1
In year 3, the value will be 80% of the previous year's value: V3 = 0.8 * V2
In year 4, the value will be 80% of the previous year's value: V4 = 0.8 * V3

Generalizing this pattern, we can define the formula as:

Vn = 0.8 * Vn-1

b.) Now, we need to find in which year the value of the bulldozer will be less than $100,000.

We can plug the formula into an equation, and solve for n:

Vn = 0.8 * Vn-1
$100,000 = 0.8 * Vn-1

Dividing both sides of the equation by $160,000, we get:

0.625 = 0.8^n-1

Taking the logarithm (base 0.8) of both sides:

log0.8(0.625) = n-1

Using a calculator, we find:

n - 1 = 4.628

Adding 1 to both sides of the equation, we get:

n = 5.628

Therefore, in the year 6 (n = 6), the value of the bulldozer will be less than $100,000.

To solve this problem, we need to find a formula to represent the value of the bulldozer each year and then use that formula to determine in which year the value will be less than $100,000.

a) To find the formula, we know that the value of the bulldozer depreciates by 20% each year. This means that the value at year n, Vn, is equal to 80% (100% - 20%) of the value at year n-1, Vn-1.

We can use this information to write a recursive formula for the value each year:
V1 = $160,000 (initial value)
Vn = 80% * Vn-1 for n > 1.

To find a more compact formula for Vn, we can expand it using the given information:
V2 = 80% * V1 = 0.8 * $160,000
V3 = 80% * V2 = 0.8^2 * $160,000
V4 = 80% * V3 = 0.8^3 * $160,000
...
Vn = 0.8^(n-1) * $160,000

So the formula for the value of the bulldozer at year n is Vn = 0.8^(n-1) * $160,000.

b) Now, we need to determine in which year the value of the bulldozer will be less than $100,000.

We can plug in the value $100,000 into the formula and solve for n:
$100,000 = 0.8^(n-1) * $160,000

To simplify the equation, let's divide both sides by $160,000:
0.625 = 0.8^(n-1)

Now, we can take the logarithm of both sides to solve for n. Using the logarithm base 0.8 (or you can also use the natural logarithm base e), we have:
Log base 0.8 (0.625) = n - 1

Using a calculator, we find that Log base 0.8 (0.625) ≈ 4.482.

So, n - 1 ≈ 4.482, which implies n ≈ 5.482.

Since n represents the year, the value of the bulldozer will be less than $100,000 in approximately year 5.482. This means that it will happen in the 6th year.

Therefore, in the 6th year, the value of the bulldozer will be less than $100,000.