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(x-2)is greater than 2(2x-1)(x-1)... please show work

Mathematics
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ok i will help
thanks
The argument of a logarithm function must always be positive, so we have the restrictions x > 0 2x - 1 > 0, equivalent to x > 1/2 x - 2 > 0, equivalent to x > 2. This is the most restrictive condition, so it governs. Taking the anti-log of both sides of the equation, we have x = (2x-1)/(x-2) x(x-2) = 2x-1 (multiply both sides by x-2) x^2 - 2x = 2x - 1 (use the distributive property to eliminate parentheses) x^2 - 4x + 1 = 0 (subtract 2x-1 from both sides to put the equation into standard form) x = (-(-4) ±√((-4)^2 - 4(1)(1)))/(2(1)) (apply the quadratic formula) x = (4 ±√(16-4))/2 (simplify) x = (4 ±2√3)/2 (simplify) x = 2±√3 (simplify) Since we know that x must be greater than 2, only the positive sign will give a suitable solution. x = 2+√3 ≈ 3.732 Check Ln[3.732] = Ln[2*3.732-1] - Ln[3.732 - 2] 1.317 = Ln[7.464 - 1] - Ln[1.732] 1.317 = Ln[6.464] - .549 1.317 = 1.866 - .549 (yes)

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is that what you r looking for
in the back of the book it says that x>1
ummm that is strang
sorry i mean that x is less than 1. it got the answer though thanks for the help
(x+1)(x+2) is greater than (x+1)??????
anyone help
(x-2) > 2(2x-1)(x-1) x-2 > 4x^2 - 6x +2
0 > 4x^2 -5x + 4
Use the AC method to factor.

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