Directions: use the properties of logarithms to write the expression as a sum, difference, or multiples of logarithms of x, y, or constant.
log (4th root of (x/y))
thanks for your help. i appreciate it
okay here is a list of log problems so that i can check it later tonight.
Use the properties of logarithms to write the expression as a sum, difference, or multiples of logarithms of x, y, constants.
1. log1000x4 <-(x to the 4th. i dont have those programs to write perfect math problems, sorry.)
2. In(3rd root of x) / (3rd root of y)
3. 1/3logx
4. 4log(x-1) + 2log(x+4)
5. 1/2log a(x-3) - 1/3log a(x+3)
6. 1/4log b(x+2) - 1/5log b(3-x)
Hi Jimgotkp! I helped with 2,3,4 and half of 5. You all check and make sure I didn't make any mistakes because I have to get off the comp.
Use the properties of logarithms to write the expression as a sum, difference, or multiples of logarithms of x, y, constants.Since $\displaystyle log_{b}a\,-\,log_{b}c\,=\,log_{b}\frac{a}{c}$, Problem # 2 can be rewritten as $\displaystyle ln{\sqrt[3]{x}}\,-\,ln{\sqrt[3]{y}}$2. $\displaystyle \ln \frac{{\sqrt[3]{x}}}{{\sqrt[3]{y}}}$
A rule says: $\displaystyle x\,log_{b}\,a\,=\,log_{b}\,a^{x}$3. $\displaystyle \frac{1}{3}logx$
so you can rewrite this is as: $\displaystyle log\,x^{\frac{1}{3}}$. I don't see how you can simplify this further.
Using the previous rule, we can rewrite this as: $\displaystyle log(x\,-\,1)^{4}\,+\,log(\,x\,+\,4)^{2}$4. $\displaystyle 4\,log(x\,-\,1)\,+\,2\,log(\,x\,+\,4)$
Then as a multiple: $\displaystyle \log [(x - 1)^4 (x + 4)^2]$
Move the fraction: $\displaystyle log_{a}(x\,-\,3)^{\frac{1}{2}}\,-\,log_{a}(x\,+\,3)^{\frac{1}{3}}$5. $\displaystyle \frac{1}{2}log_{a}(x\,-\,3)\,-\,\frac{1}{3}log_{a}(x\,+\,3)$
Once again,..$\displaystyle log_{b}a\,-\,log_{b}c\,=\,log_{b}\frac{a}{c}$: .... $\displaystyle log_{a}\frac{(x\,-\,3)^{\frac{1}{2}}}{(x\,+\,3)^{\frac{1}{3}}}$
lol, i'm not a moderator!
are you sure these are the instructions for all the questions? some of these are already in the form you are asking us to put them in, it would be a better exercise to combine them rather than express them as sums and whatnot. i'll do the ones Jonboy left out.
$\displaystyle \log 1000x^4 = \log 1000 + \log x^4 = 3 + 4 \log x$ .......i assumed we are dealing with log to the base 10 here, so that's why i got the 3
$\displaystyle \frac {1}{4} \log_{b} (x + 2) - \frac {1}{5} \log_{b} (3 - x) = \log_{b} (x + 2)^{ \frac {1}{4}} - \log_{b} (3 - x)^{ \frac {1}{5}}$6. 1/4log b(x+2) - 1/5log b(3-x)
$\displaystyle = \log_{b} \sqrt [4] {x + 2} - \log_{b} \sqrt [5] {3 - x}$
$\displaystyle = \log_{b} \left( \sqrt [4] {x + 2} \sqrt [5] {3 - x} \right)$
if you have any questions, please feel free to ask