exponential log 22

Question

A computer purchased for $550 loses 17% of its value every year.The computer's value can be modeled by the function , where  is the dollar value and  the number of years since purchase.(A) In the exponential model   and  .

(B) In how many years will the computer be worth half its original value? 

Round answer to 1 decimal place.The answer is____  years?

Answer 1

For all exponential functions - either growth (gain, such as interest) or decay (loss, such as depreciation - as is this case), there is a starting point and a growth/decay rate. These values are part of the standard exponential equation.

a) In v(t) = a ∗ b^t, 'a' represents the starting point and 'b' represents the growth/decay rate. In this case...

a = 550

b = (1 - 0.17) = 0.83

b) To find how many years it takes to reach a certain value, substitute what you know first then apply logarithms to determine the time. In this case...

275 = 550 ∗ (0.83)^t

(Dividing by 550) 0.5 = (0.83)^t

(Using logarithms) log (0.5) = t log (0.83)

(Solve for 't') t = log (0.5)/ log (0.83)

≈ -0.3010 / -0.089

≈ 3.7 years

Answer 2

a = 550

b = 0.083

0.5 = (0.83)^t

ln 0.5 = ln 0.83 t

-0.6931 ≈ -0.1863 t

3.7170 ≈ t

Answer 3

(A) v(t) = ab^t; if t = 0 then v(0) = ab⁰ = a = 550.Now v(t) = 550b^t.

To mget b we set t = 1 year then v(1) = 550 - 550·017 = 456.5. So 456.5 = 550b, b = 456.5/550 = 0.83

Auer exponential model v(t) = 550·0.83^t.

a = $550, b = 0.83

(B) Here v(t) = 1/2·550·= $275

To get time t we have equation: 550·0.83^t = 275, or 0.83^t = 0.5, t = ln(0.5)/ln(0.83) = 3.72 years