write as a single logarithm:
ln(x/x-1)+ ln(x+1/x)- ln(x^2-1)
please show me the steps.
thanx
Michelle, I hope you know the three main rules of logs.
1. log (AB) = log A + log B
2. log (A/B) = log A - log B
3 log A^n - n log A
Using these rules and guessing at your order of operations you would have
ln[{x/(x-1)}{(x+1)/x}all over (x^2-1)]
factoring the bottom into (x+1)(x-1) and reducing the inside of the square bracket I got
ln(1/(x-1)^2)
now using rule #2 and #3 you would get
ln 1 - ln(x-1)^2
= 0-2ln(x-1)
=-2ln(x-1)
thank u sooooo much. been tryin to figure out wat ive been doing wrong.