If the inequality is true, determine x? log 1/4 (x^2 - 6x + 6)>=0

Expert Answers
hala718 eNotes educator| Certified Educator

log 1/4 (x^2 - 6x+  6) >= 0

We know that in order for log to be positive, then :

x^2 - 6x + 6 >= (1/4)^0

==> x^2 - 6x +6 >= 1

Now subtract 1 from both sides:

==> x^2 - 6x + 5 >= 0

Now factor:

== (x -5)(x-1) >=0

==> (x-5)>=0    AND    x-1>=0

==> x>=5   AND  x >= 1

==> x = [5, inf)

ALso:

(x-5) =< 0      AND   (x-1) =< 0

==> x =< 5    AND    x =< 1

==> x = (-inf, 1]

Then x= (-inf, 1] U [5, inf)

 

neela | Student

To check whether log1/4(x^2-6x+6) > 0.

 

So the inequality, log (1/4) (x^2-6x+6) > = 0 we pressume to be true and find conditions if any for it to be true.

Taking anti log, we get:

(1/4)(x^2-6x+6) > = 1, as log (1) = 0

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

x^2-6x+5 > = 0

(x-5)(x-1) >=0.

Or x <1  or  x>5, when this product is positive,

So x should be < 1 Or x > 5 in order that the inequality holds.

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giorgiana1976 | Student

Because the logarithm has a subunit base, namely 1/4, the logarithmic function is decreasing.

We'll re-write the equation:

 log 1/4 (x^2 - 6x + 6) >= 0

 log 1/4 (x^2 - 6x + 6) >= log 1/4 1

Because the bases are matching, we'll apply one to one property and we'll change the direction of the inequality:

x^2 - 6x + 6 =< 1

We'll subtract 1 both sides:

x^2 - 6x + 6 - 1=< 0

x^2 - 6x + 5 =< 0

We'll calculate the roots of the equation:

x^2 - 6x + 5 = 0

We'll apply the quadratic formula:

x1 = [6 + sqrt(36-20)] / 2

x1 = (6+4) / 2

x1 = 5

x2 = (6-4) / 2

x2 = 1

The expression is negative over the interval [1 , 5] and is positive for x< 1 and x> 5.