# Re: Algebra II

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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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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?