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Re: Algebra II

posted by on .

How to solve
Log (9x+5) - Log ((x^2)-1) = 1/2
16 16

(The 16 is from the logarithmic function : b) (y=log x)
b

Could anyone post the answer to this?

16 is supposed to be next to Log(16) but lower. This is for both the Logs in the equation.

  • Re: Algebra II - ,

    So we are looking at
    log16 (9x+5) - log16</sub (x^2 - 1) = 1/2
    log16</sub [(9x+5)/(x^2-1)] = 1/2
    (9x+5)/(x^2-1) = 16^(1/2)
    (9x+5)/(x^2-1) = 4
    4x^2 - 4 = 9x + 5
    4x^2 - 9x - 9 = 0
    (x-3)(4x + 3) = 0
    x = 3 or x = -3/4


    but when x = -3/4, the first log term is undefined (we can't take a log of a negative)

    so x = 3

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