# expoenential

• Mar 25th 2006, 02:08 AM
rachael
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.
• Mar 25th 2006, 04:18 AM
earboth
Quote:

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.

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