Homework Help

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

user profile pic

lixalixa | (Level 2) Honors

Posted September 4, 2013 at 4:26 PM via web

dislike 0 like

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

1 Answer | Add Yours

user profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted September 4, 2013 at 4:51 PM (Answer #1)

dislike 0 like

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.

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes