Question

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

Answer 1

ok i will help

Answer 2

thanks

Answer 3

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)

Answer 4

is that what you r looking for

Answer 5

in the back of the book it says that x>1

Answer 6

ummm that is strang

Answer 7

sorry i mean that x is less than 1. it got the answer though thanks for the help

Answer 8

(x+1)(x+2) is greater than (x+1)??????

Answer 9

anyone help

Answer 10

(x-2) > 2(2x-1)(x-1)
x-2 > 4x^2 - 6x +2

Answer 11

0 > 4x^2 -5x + 4

Answer 12

Use the AC method to factor.