Asked by Anonymous
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.
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.
Answers
Answered by
Reiny
So we are looking at
log<sub>16</sub> (9x+5) - log<sub>16</sub (x^2 - 1) = 1/2
log<sub>16</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
log<sub>16</sub> (9x+5) - log<sub>16</sub (x^2 - 1) = 1/2
log<sub>16</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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