Question

The length (in centimeters) of a typical Pacific halibut t years old is approximately f(t) = 200(1 − 0.956e−0.18t). Suppose a Pacific halibut caught by Mike measures 148 cm. What is its approximate age? (Round your answer to one decimal place.)

Answer 1

Its approximate age is 7.2 years.

We will find the approximate age by solving the equation

f(t) = 200 (1-0.956e^(0.18)....(eq.1)

Since f(t) = 148cm

f(t) = 148cm, because pacific halibut caught by Mike measures 148cm.

Now put f(t) = 148 and solve the equation (1) for t:

148 = 200(1 - 0.956e ^ (- 0.18t))

148 = 200 - 191.2e ^ (- 0.18t)

191.2e ^ (- 0.18t) = 200 - 148

191.2e ^ (- 0.18t) = 52

e ^ (- 0.18t) = 52/191.2

e ^ (- 0.18t) = 0.271966527

- 0.18t = ln(0.271966527)

- 0.18t = - 1.30207628

t = 1.30207628/0.18

t≈ 7.2 years

Therefore, its approximate age is 7.2 years.

Answer 2

7.2 years

Explanation:

`f(t) =200(1 - 0.956e^(-0.18t))`

where f(t) is the length and t is the number of years old

given the Mike measures 148 cm, we need to find out the age

So we plug in 148 for f(t) and solve for t

`148 =200(1 - 0.956e^(-0.18t))`

divide both sides by 200

`(148)/(200) = 1 - 0.956e^(-0.18t)`

Now subtract 1 from both sides

`-0.26= - 0.956e^(-0.18t)`

Divide both sides by -0.956

`(0.26)/(0.956) =e^(-0.18t)`

Now take ln on both sides

`(0.26)/(0.956) =e^(-0.18t)`

`ln((0.26)/(0.956) )=-0.18tln(e)`

`ln((0.26)/(0.956) )=-0.18t`

divide both sides by -0.18

t=7.2

So 7.2 years