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Log(base6) (x+3) + log(base 6) (x+4) =1 solve equation for exact value of x. please help?

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3 months ago 5

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  1. Outlier

    log_6 (x + 3) + log_6 (x + 4) = 1

    Note these rules:
    1) log (ab) = log(a) + log(b)
    2) log_a (a) = 1

    So we can rewrite the equation as:

    log_6 [(x + 3)(x + 4)] = log_6 (6)

    We can take the anti-log of both sides to convert this equation to one not involving logarithms:

    (x + 3) (x + 4) = 6

    Expand brackets, simplify and factorize:
    x^2 + 4x + 3x + 12 = 6
    x^2 + 7x + 12 = 6
    x^2 + 7x + 6 = 0
    x^2 + 6x + x + 6 = 0
    x(x + 6) + 1(x + 6) = 0
    (x + 6)(x + 1) = 0

    x = -6 or x = -1

    But note that x = -6 cannot be used as a solution because it is undefined for the terms log_6 (x + 3) and log_6 (x + 4).
    That is, if x = -6, we get log_6 (-3) and log_6 (-2) and these values are undefined.

    So the solution is x = -1

  2. Rogue

    log₆(x + 3) + log₆(x + 4) = 1
    given log(a) + log(b) = log(ab)
    ∴ log₆((x + 3)(x + 4)) = 1
    ∴ 6^log₆((x + 3)(x + 4)) = 6¹
    given a^log_a(n) = n
    ∴ (x + 3)(x + 4) = 6
    ∴ x² + 7x + 12 = 6
    ∴ x² + 7x + 6 = 0
    ∴ (x + 1)(x + 6) = 0
    ∴ x = -1 and x = -6
    given x = -6 is outside of domain of the original question
    ∴ x = -1

  3. TomV

    log₆(x+3) + log₆(x+4) = 1 : x >-3
    log₆[(x+3)(x+4)] = 1
    (x+3)(x+4) = 6
    x² + 7x + 6 = 0
    (x+6)(x+1) = 0
    x = -6, -1
    Discard the x = -6 result as x must be greater than -3 for the equation to be real valued.

    x = -1

  4. Iggy Rocko

    log(base 6)(x + 3) + log(base 6)(x + 4) = 1
    log(base 6)((x + 3)(x + 4)) = 1
    log(base 6)(x^2 + 7x + 12) = 1
    6^1 = x^2 + 7x + 12
    0 = x^2 + 7x + 6
    (x + 6)(x + 1) = 0
    x = -6 or x = -1
    Discard x = -6 because it leads to imaginary numbers , leaving just
    x = -1

  5. alex

    (x+3)(x+4)=6 , solve for x , with x >-3

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