Posted by Anonymous on Tuesday, May 18, 2010 at 5:21pm.
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  Reiny, Tuesday, May 18, 2010 at 5:33pm
So we are looking at
log_{16} (9x+5)  log_{16</sub (x^2  1) = 1/2
log16</sub [(9x+5)/(x^21)] = 1/2
(9x+5)/(x^21) = 16^(1/2)
(9x+5)/(x^21) = 4
4x^2  4 = 9x + 5
4x^2  9x  9 = 0
(x3)(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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