Re: Algebra II
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.

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}