10. A bacteria is growing continually at a rate of 2.34% per hour. If the bacteria is first foundto be 850mg, in how many hours will it take to reach 1000mg?
Accepted Solution
A:
Answer:The time to reach 1000 mg is 7 hoursStep-by-step explanation:Given as :The rate at which bacteria growing = 2.34% per hourThe initial amount of bacteria = 850 mgThe final amount of bacteria = 1000 mgLet The time period fro bacteria growing = T hoursNow, Final value = Initial value × [tex](1 + \frac{Rate}{100})^{Time}[/tex]Or, 1000 mg = 850 mg × [tex](1 + \frac{2.34}{100})^{T}[/tex]Or, [tex]\frac{1000}{850}[/tex] = [tex](1 + \frac{2.34}{100})^{T}[/tex] Or, 1.1764 = [tex](1 + \frac{2.34}{100})^{T}[/tex] Or, 1.1764 = [tex](1.0234)^{T}[/tex] Or, Taking log both side log(1.1764) = log ( [tex](1.0234)^{T}[/tex] )Or, 0.07055 = T × 0.01004 ( [tex]loga^{b} = b log a[/tex])∴ T = 7.026 hour or T ≈ 7 hoursHence The time to reach 1000 mg is 7 hours Answer