A laptop computer is purchased for $4800. Each year, its value is 80% of its value the year before. After how many years will the laptop computer be worth 2000 or less?

Answer 1

Hi there,

This is a hidden geometric sequence problem considering that the next value is always being multiplied by the same number, 0.8:

The key equation is: a_n=a₁rⁿ (here we're taking the first year to start at zero and not one)

Where: a₁=4800

r=0.8

and a_n=2000

So,

2000=4800(0.8)ⁿ

Solve for "n" algebraically.

5/12=(0.8)ⁿ

n= log(5/12)/log(0.8)= 3.9 years,

So 4 years.

I hope this helps!