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another pre cal (logs)

posted by on .

write as a single log

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

thanks

  • another pre cal (logs) - ,

    according to your basic laws of logs your expression is

    log3 ((3x-6)(x^2 - 4)/81)

  • sorry, try again - ,

    I misread you + and - signs

    should be

    log3 (81(3x-6)/(x^2 - 4))

  • sorry, try again - ,

    which reduces to

    log3 243/(x-2) , x > 2

  • another pre cal (logs) - ,

    Thank you

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