Exponential problem with E. coli

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The discussion centers on a mathematical model for the concentration of E. coli in a stream over time, given by the equation x(t) = loge(t + e^2). Participants clarify that to find the initial concentration of E. coli, one should evaluate the function at t=0, which results in 2 parts per million (ppm). The introduction of E. coli occurs at time t=0. The conversation confirms the correct interpretation of logarithmic functions in this context. Overall, the problem illustrates the application of logarithmic equations in environmental science.
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20 The number of parts per million, x, of E. coli in a stream t hours after a pollutant containing E. coli is introduced is modeled by
x(t) = loge (t + e2), t ≥ 0.
a How many parts per million of E. coli are introduced into the stream?
how do i work out this question?
 
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At what time t does the introduction of E coli take place?

Regards,
George
 
I'm assuming you know the laws of logs? How far have you worked through the problem?
 
thank you . ...
 
i let t=0 therefore loge e2 equals to 2 is it correct?
 
rachael said:
i let t=0 therefore loge e2 equals to 2 is it correct?

2 ppm, yes. Nice work!

Regards,

Rich B.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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