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a cup of coffee contains 140 mg of caffeine. if caffeine leaves the body at 10% per hour, how long will it takes for half of the caffeine to be eliminated from ones body

User U And Me
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Explanation:

10% per hour is tricky, because for the first hour it is 10% of the original 140 mg (14 mg). but for the second hour it is 10% of the updated level of 140 - 14 = 126 mg (12.6 mg). and so on.

so, we cannot simply say 5× 10% = 50%, therefore, it will take 5 hours.

no, instead we must use arithmetic sequences.

10% is represented as mathematical factor as 0.1.

so, when 10% are removed, it means that 90% remain.

90% = 0.9.

90% of 100% = 100% × 0.9

so,

a1 = 140

a2 = a1×0.9

a3 = a2×0.9 = a1×0.9²

an = a1×0.9^(n-1)

we need to find n for which an = 70 (half of the caffeine remains, meaning half of the caffeine is gone).

70 = 140×0.9^(n-1)

0.5 = 0.9^(n-1)

using the logarithm to the base of 0.9 to solve :

log0.9(0.5) = n - 1

n = log0.9(0.5) + 1

how to get log0.9(0.5) ?

all logarithms are related to each other :

loga(b) = logc(b)/logc(a)

in our case we can use ln (base e) or log (base 10) - whatever your calculator offers :

log0.9(0.5) = log(0.5)/log(0.9) = 6.578813479...

n = 6.578813479... + 1 = 7.578813479... hours

0.578813479...×60 = 34.72880874... minutes

so, after about 7.6 hours or 7 hours and 35 minutes half of the caffeine will be gone.

User Klutt
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