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precalculus

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log5 (x+3)= 1- log5 (X-1) where do I start?

  • precalculus -

    start by collecting logs:

    log5 (x+3) + log5(x-1) = 1

    now recall that sum of logs is log of product

    log5 [(x+3)(x-1)] = 1

    raise 5 to the power of both sides:

    (x+3)(x-1) = 5
    x^2 + 2x - 3 = 5
    x^2 + 2x - 8 = 0
    (x+4)(x-2) = 0

    x = 2 or -4

    Since logs of negative numbers do not exist, we have to throw out x = -4 since it does not fit the original equation.

    x=2 is our only solution.

    check:
    log5 (2+3) = 1 - log5 (2-1)
    log5 5 = 1 - log5 1
    1 = 1 - 0

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