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March 29, 2015

March 29, 2015

Posted by **Deb** on Monday, March 31, 2008 at 1:14am.

log(underscore3)(3x-6)-[log(underscore3)(x^2-4)+log(underscore3)81]

thanks

- another pre cal (logs) -
**Reiny**, Monday, March 31, 2008 at 1:17amaccording to your basic laws of logs your expression is

log_{3}((3x-6)(x^2 - 4)/81)

- sorry, try again -
**Reiny**, Monday, March 31, 2008 at 1:19amI misread you + and - signs

should be

log_{3}(81(3x-6)/(x^2 - 4))

- sorry, try again -
**Reiny**, Monday, March 31, 2008 at 1:22amwhich reduces to

log_{3}243/(x-2) , x > 2

- sorry, try again -
- another pre cal (logs) -
**Deb**, Monday, March 31, 2008 at 1:37amThank you

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