SOLUTION: Coffee contains caffeine. The half-life of caffeine is 5 hours. This means the amount of caffeine in your bloodstream is reduced by 50% every 5 hours. Suppose you drink a cup of co
Algebra ->
Exponential-and-logarithmic-functions
-> SOLUTION: Coffee contains caffeine. The half-life of caffeine is 5 hours. This means the amount of caffeine in your bloodstream is reduced by 50% every 5 hours. Suppose you drink a cup of co
Log On
Question 1171695: Coffee contains caffeine. The half-life of caffeine is 5 hours. This means the amount of caffeine in your bloodstream is reduced by 50% every 5 hours. Suppose you drink a cup of coffee that contains 320 mg of caffeine. How long will it take until there is 5 mg of caffeine left in your bloodstream? Found 2 solutions by Boreal, math_tutor2020:Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! the half-life is 5 hours.
5/320=1/64 which is 2^-6, or (1/2)^6
(1/2)^6=1/64
6 half-lives or 30 hours.
320-160(5)-80(10)-40(15)-20(20)-10(25)-5(30)
You can put this solution on YOUR website!
The phrasing "is reduced by 50%" is the same as "reduced by half". Every 5 hours, the amount of caffeine is reduced by half.
If you started with 320 mg of caffeine, then you'll have 320/2 = 160 mg of caffeine after 5 hours. Then you'll have 160/2 = 80 mg after another 5 hours (5+5 = 10 total so far)
We keep dividing the amount in half and keep track how much time has passed by (along with the number of half-lives)
A table is useful for this sort of thing.
Half-life
Time
Amount
0
0
320
1
5
160
2
10
80
3
15
40
4
20
20
5
25
10
6
30
5
The time value is in hours, and the amount represents the amount of caffeine in mg.
The table shows at time 30 hours, there's exactly 5 mg of caffeine left in the bloodstream.
This happens after 6 half-lives have occurred.
----------------------------------------------
An algebraic approach
x = number of half-lives that pass by
y = amount of caffeine left over in the bloodstream
The equation tying together x and y is
y = 320*(0.5)^x
where 1/2 = 0.5 represents the multiplier indicating a 50% reduction
We could guess-and-check by plugging in random values of x to see if we get y = 5.
But let's solve for x using algebra
Plug in y = 5 and use logs to isolate the exponent
y = 320*(0.5)^x
5 = 320*(0.5)^x
5/320 = (0.5)^x
0.015625 = (0.5)^x
log(0.015625) = log( (0.5)^x )
log(0.015625) = x*log( 0.5 )
x*log( 0.5 ) = log(0.015625)
x = log(0.015625)/log(0.5)
x = 6
This helps confirm the half-life we got earlier with the table.
Another confirmation method is to plug x = 6 into the equation
y = 320*(0.5)^x
y = 320*(0.5)^6
y = 320*0.015625
y = 5
x = 6 half-lives occurring mean 5x = 5*6 = 30 hours have passed by (since each half-life is 5 hours).
(