Question

10. A bacteria is growing continually at a rate of 2.34% per hour. If the bacteria is first found

to be 850mg, in how many hours will it take to reach 1000mg?​

Answer 1

The time to reach 1000 mg is 7 hours

Explanation:

Given as :

The rate at which bacteria growing = 2.34% per hour

The initial amount of bacteria = 850 mg

The final amount of bacteria = 1000 mg

Let The time period fro bacteria growing = T hours

Now, Final value = Initial value ×
`(1 + (Rate)/(100))^(Time)`

Or, 1000 mg = 850 mg ×
`(1 + (2.34)/(100))^(T)`

Or,
`(1000)/(850)` =
`(1 + (2.34)/(100))^(T)`

Or, 1.1764 =
`(1 + (2.34)/(100))^(T)`

Or, 1.1764 =
`(1.0234)^(T)`

Or, Taking log both side

log(1.1764) = log (
`(1.0234)^(T)` )

Or, 0.07055 = T × 0.01004 (
`loga^(b) = b log a`)

∴ T = 7.026 hour or T ≈ 7 hours

Hence The time to reach 1000 mg is 7 hours Answer