Question

Timothy purchased a computer for $1,000.

Answer 1

Answer:

This situation represents [a.) exponential decay

The rate of growth or decay, r, is equal to  c.) 0.2

So the value of the computer each year is g.) 80% of the value in the previous year.

It will take [i.) 3 years for the value of the computer to reach $512.

Step-by-step explanation:

Since the value depreciates each year, it represents exponential decay.

The rate of growth or decay, r, is equal to

Depreciates 20%, so r = 0.2.

So the value of the computer each year is ... of the value in the previous year,

Depreciates 20%, so it is 100% - 20% = 80% of the value in the previous year.

It will take x years for the value of the computer to reach $512.

The value of the computer after x years is given by:

$V(x) = V(0)(1-r)^x$

In which V(0) is the initial value and r is the decay rate. So

$V(x) = 1000(1-0.2)^x$

$V(x) = 1000(0.8)^x$

We want to find x for which V(x) = 512. So

$V(x) = 1000(0.8)^x$

$512 = 1000(0.8)^x$

$(0.8)^x = \frac{512}{1000}$

$(0.8)^x = (0.512)$

Applying log to both sides:

$\log{(0.8)^x} = \log{(0.512)}$

$x\log{0.8} = \log{(0.512)}$

$x = \frac{\log{0.512}}{\log{0.8}}$

$x = 3$

So it will take 3 years.