Posted by Anonymous 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 
Reiny,
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}