A laptop computer is purchased for $2300. Each year, its value is 80% of its value the year before. After how many years will the laptop computer be worth $800 or less?

To solve this problem, we need to understand that the value of the laptop decreases by 20% each year. This is a geometric sequence problem where the initial term (a) is $2300, the common ratio (r) is 0.8 (which is 80$800 or less.

The formula for the nth term of a geometric sequence is:

an = a * r^(n-1)

We can plug the values into this formula and solve for n:

800 = 2300 * (0.8)^(n-1)

To solve for n, we first divide both sides by 2300:

0.3478 = (0.8)^(n-1)

Then we take the natural logarithm (ln) of both sides:

ln(0.3478) = ln((0.8)^(n-1))

Using the property of logarithms that allows us to bring the exponent out front, we get:

ln(0.3478) = (n-1) * ln(0.8)

Finally, we solve for n:

n = [ln(0.3478) / ln(0.8)] + 1

Using a calculator, we find that n is approximately 6.3. Since we can't have a fraction of a year, we round up to the nearest whole number.

So, it will take 7 years for the laptop to be worth $800 or less.