Question

Darrin drank a latte with 205 milligrams (mg) of caffeine. Each hour, the caffeine in Darrin’s body diminishes by about

Answer 1

Answer:

a. C(t)=205*(1-0.08)^t

b. t=log_0.92(C(t)/205)=(log_10(C(t)/205))/(log_10(0.92))

c. 16.92 hours

Step-by-step explanation:

Let's say that C(t) is the expression of the amount of caffeine remaining in Darrin's system after t time, hours in this particular case.

a. Then for the first hour the expression would be:

C(t)=205*(1-0.08)

For the second hour:

C(t)=205*(1-0.08)-205*(1-0.08)*(1-0.08)

For the third

C(t)=205*(1-0.08)-205*(1-0.08)*(1-0.08)-205*(1-0.08)*(1-0.08)*(1-0.08)

And so on, for that reason the best way to fit the expression is:

C(t)=205*(1-0.08)^t

2. To find the correct expression for time, we must solve for t the equation recently written above:

Considering that  log_b(a)=c  and  log_b⁡(a)=log_c⁡(a)/log_c⁡(b), then:

t=log_0.92(C(t)/205)

t= (log_10(C(t)/205))/(log_10(0.92))

3. Finally we replace the given value of C(t) into the equation for t:

t= (log_10(50/205))/(log_10(0.92))=16.92

t= 16.92 hours