Question

Timothy purchased a computer for $1,000.


The value of the computer depreciates by 20% every year.


This situation represents [a.) exponential decay, b.) exponential growth]


The rate of growth or decay, r, is equal to [c.) 0.8, d.) 1.2, e.)0.2

year.


So the value of the computer each year is [f.) 20 %, g.) 80%, h.) 120%] of the value in the previous year.


It will take [i.) 3, j.) 2, k.) 4, l.) 5.] years for the value of the computer to reach $512.

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.