expoenential

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 modelled by x(t) = loge (t + e^2), t ≥ 0.
a How many parts per million of E. coli are introduced into the stream?
b To the nearest tenth of a part per million, how many parts per million of E. coli are in the stream when t = 10?


could some one help me with part b?
i tried subin t=10
therefore i got x(10)= loge ( 10 + e^2)
= loge 10 * loge e^2
= 4.605
but the answer is wrong.

Quote:

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Originally Posted by rachael

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 modelled by x(t) = loge (t + e^2), t ≥ 0.
a How many parts per million of E. coli are introduced into the stream?
b To the nearest tenth of a part per million, how many parts per million of E. coli are in the stream when t = 10?

could some one help me with part b?
i tried subin t=10
therefore i got x(10)= loge ( 10 + e^2)
= loge 10 * loge e^2
= 4.605
but the answer is wrong.

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Hello,

you only have to plug in the values you know into the given equation:
$\displaystyle x(t) = log_{e} (t + e^2)$. If t = 10, you'll get:
$\displaystyle x(t) = log_{e} (10 + e^2)\approx 2.8558...$

(By the way: You are right: The logarithm of a product is a sum, but the logarithm of a sum is not a product.)

(For $\displaystyle log_{e}$ you better use $\displaystyle ln$ and then you'll find the correct button on your calculator more easily.)

Greetings

EB