A marine biologist measures the presence of a pollutant in an ocean and concludes that the concentration, C, in parts per million (ppm), as a function of the population, P, of the people who visit the beach is given by C(P)=1.38P+ 97.4 . The population of people visiting the beach, in thousands, can be modeled by P(t)=12(1.078)^t where t is the time in years since the first measurement.

 

 

a) Determine an equation, in simplified form, for the concentration of pollutant as a function of the number of years since the first measurement.

 

b) How long (to the nearest year) will it take for the concentration to reach 180 ppm?

a) ?C(P(t))=1.38[12(1.078)t]+97.4?

b) ?t≈21? years

We are given the function for concentration of pollutants in parts per million ?C(P)=1.38P+97.4?

On the other hand, the function for the population of people visiting the beach in thousands is modeled by ?P(t)=12(1.078)t?

where:

?t? is the time in years since the first measurement.

a) Determine an equation, in simplified form, for the concentration of pollutant as a function of the number of years since the first measurement.

?C(P(t))=1.38[12(1.078)t]+97.4?

b) How long (to the nearest year) will it take for the concentration to reach 180 ppm?

So, we have,

?180=1.38[12(1.078)t]+97.4?

Solving for t, we have,

Switching sides, we have,

?1.38(12(1.078)t)+97.4=180?

Simplifying, we have,

?16.56⋅1.078t+97.4=180?

Multiplying both sides by 100, we have,

?16.56⋅1.078t⋅100+97.4⋅100=180⋅100?

?1656⋅1.078t+9740=18000?

Subtracting 9740 from both sides, we have,

?1656⋅1.078t+9740−9740=18000−9740?

?1656⋅1.078t=8260?

Dividing both sides by 1656, we have,

?16561656⋅1.078t?=16568260??

?1.078t=4142065??

Using the rule: ?Iff(x)=g(x),thenln(f(x))=ln(g(x))? , we have,

?ln(1.078t)=ln(4142065?)?

Applying the log rule: ?loga?(xb)=b⋅loga?(x)? , we have,

?tln(1.078)=ln(4142065?)?

So, we have,

?t=ln(1.078)ln(4142065?)??

?t=21.39626695? years or

?t≈21? years