take a look at the post by Krystal (unless you are that same person, the posting looks very similar)
http://www.jiskha.com/display.cgi?id=1357052002
all you have to change is the base from 1/3 to 1/6
I think you can handle the rest.
solve the following equation.
log 1/6 (x^2+x)- log 1/6 (x^2-x)=2
**the 1/6 is lowered on both sides and the (x^2+x) and (x^2-x) should be higher on both sides**
3 answers
(x^2+x)/(x^2-x) = (1/6)^2
(x+1)/(x-1) = 1/36
36x+36 = x-1
35x = -37
x = -37/35
(x+1)/(x-1) = 1/36
36x+36 = x-1
35x = -37
x = -37/35
Thanks Steve. I don't know why but when I see (1/6)^2 for some reason I want to just multiply the 6*2. It for some reason doesn't seem to stick that it is 6*6 to equal 36. I worked the problem--just had different numbers since I multiplied 6*2 to get 12.
18 answers