1. ## URGENT HELP!!! pre-test

questions are on attachment!!!

2. #1 A. Use the property that $\displaystyle \log(a)-\log(b)=\log\frac{a}{b}$

3. Hello, Brooke!

Am I reading the problem wrong?
I don't get any of their answers . . .

2. The number of bison per acre of range in the wild is: $\displaystyle N \:=\:2.1 \times 10^{5-0.004w}$
where $\displaystyle N$ is the number of bison per acre
and $\displaystyle w$ is the average weight of the bison in pounds.

Find the average weight of a bison in a herd that has an average
of seven animals per acre of range.

. . A: 1247 lbs . . B: 1084 lbs . . C: 1134 lbs . . D: 1020 lbs

We are given: $\displaystyle N = 7$

So we have: .$\displaystyle 2.1 \times 10^{5-0.004w}\:=\:7\quad\Rightarrow\quad 10^{5-0.004w}\:=\:\frac{10}{3}$

Then: .$\displaystyle 5-0.004w\:=\:\log\left(\frac{10}{3}\right)\quad \Rightarrow\quad 0.004w\:=\:5 - \log\left(\frac{10}{3}\right)$

Therefore: .$\displaystyle w \;= \;\frac{5 - \log\left(\frac{10}{3}\right)}{0.004}\;=\;1119.280 314$

4. Originally Posted by Jameson
#1 A. Use the property that $\displaystyle \log(a)-\log(b)=\frac{\log(a)}{\log(b)}$
I'm sorry, Jameson, but I don't think that property holds.

$\displaystyle log(a)-log(b)=log(\frac{a}{b})$

$\displaystyle log(a)-log(b)\neq\frac{log(a)}{log(b)}$

5. Originally Posted by galactus
I'm sorry, Jameson, but I don't think that property holds.

$\displaystyle log(a)-log(b)=log(\frac{a}{b})$

$\displaystyle log(a)-log(b)\neq\frac{log(a)}{log(b)}$
You are quite right. It was a Latex typo. Thanks for spotting it.