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questions are on attachment!!!
#1 A. Use the property that $\displaystyle \log(a)-\log(b)=\log\frac{a}{b}$
Hello, Brooke!
Am I reading the problem wrong?
I don't get any of their answers . . .
Quote:
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: .$\displaystyle w \;= \;\frac{5 - \log\left(\frac{10}{3}\right)}{0.004}\;=\;1119.280 314$
I'm sorry, Jameson, but I don't think that property holds.Quote:
Originally Posted by Jameson
$\displaystyle log(a)-log(b)=log(\frac{a}{b})$
$\displaystyle log(a)-log(b)\neq\frac{log(a)}{log(b)}$
You are quite right. It was a Latex typo. Thanks for spotting it. :DQuote:
Originally Posted by galactus
