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
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