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calculus

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log (3x-9) = 2log3 - log27+ log(x+3)

Solving the euqation using the properties of logarithms

How do you do this?

  • calculus - ,

    I will assume that all logs have the same base.

    log(3x -9) = log [3^2*(x+3)/(3^3]
    = log[(x+3)/3]

    Therefore
    3x -9 = (x+3)/3
    9x -27 = x +3
    8x = 30
    x = 3.75

    Check:
    log (3x -9) = log 2.25 = 0.35218

    2log3 -log27 + log(6.75)
    = 0.95424 -1.43136 + 0.82930
    = 0.35218

    It makes no difference what log base you choose, as long as it is the same for all terms. (I chose base 10 when checking)

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