# Show that y=(1/2)^(log (base 1/9)(2x^2-3x+1)) -1 is negative

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The expression y = `(1/2)^(log_(1/9)(2x^2-3x+1)) -1` has to be shown to be negative. As `(1/2)` is positive `(1/2)^(log_(1/9) (2x^2 - 3x + 1))` is also positive.

For log_(1/9) (2x^2 - 3x + 1) to be defined 2x^2 - 3x + 1 > 0

`2x^2 - 3x + 1`

= `2x^2 - (2*sqrt 2*3)/(2*sqrt 2)*x + 1/8 + 1 - 1/8`

= `(sqrt 2*x - 1/(2*sqrt 2))^2 + 7/8`

The minimum value of `log_(1/9) (2x^2 - 3x + 1)` is 0.06 and `(1/2)^0.06` < 1

For greater values of `log_(1/9) (2x^2 - 3x + 1)` , the value of `(1/2)^(log_(1/9) (2x^2 - 3x + 1))` is smaller.

**As a result the expression `(1/2)^(log_(1/9) (2x^2 - 3x + 1)) - 1` is always negative.**