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write as a single logarithm:

ln(x/x-1)+ ln(x+1/x)- ln(x^2-1)

please show me the steps.


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


now using rule #2 and #3 you would get

ln 1 - ln(x-1)^2
= 0-2ln(x-1)

thank u sooooo much. been tryin to figure out wat ive been doing wrong.

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