Solve log 1/2 (2x^2 + x + 1) >= -1

3 Answers | Add Yours

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

log 1/2 (2x^2 + x + 1 ) >=-1

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

==> 2x^2 + x + 1 >= 2

Now subtract 2 from both sides:

==> 2x^2 + x -1 >= 0

Now let us factorise:

==> (2x -1)(x+1) >= 0

==> (2x-1) > 0    and (x+1)  >0

==> x > 1/2   and  x> -1

==> x belongs to (1/2, inf.)

Also,

(2x-1) < 0    and (x+1) <0

==> x < 1/2    and x < -1

==> x belongs to  (-inf, -1)

Then the solution is:

x belongs to (-inf, -1) U (1/2, inf)

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

log1/2 (2x^2+x+1) >= -1.

Solution:

The base of logarithm is (1/2).

Taking antilog , we get:

2x^2+x+1 > =  (1/2)^-1 = 1/(1/2) =  2

2x^2+x+1 > = 2

2x^2 +x+1-2 >= 0

2x^2 +x-1 > 0

(2x-1)(x+1) > 0

2x -1 > 0 Or x < -1.

x >1/2 Or x < -1.

Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Because the logarithm has a subunit base, namely 1/2, the logarithmic function is decreasing.

We'll re-write the equation:

 log 1/2 (2x^2 + x + 1) >= -1*1

log 1/2 (2x^2 + x + 1) >= -1*log 1/2 (1/2)

log 1/2 (2x^2 + x + 1) >= log 1/2 (1/2)^-1

log 1/2 (2x^2 + x + 1) >=  log 1/2 (2)

2x^2 + x + 1 =< 2

We'll subtract 2 both sides:

2x^2 + x + 1 - 2 =< 0

2x^2 + x - 1 =< 0

We'll calculate the roots of the equation:

2x^2 + x - 1 = 0

We'll apply the quadratic formula:

x1 = [-1 + sqrt(1 + 8)] / 4

x1 = (-1 + 3) / 4

x1 = 2/4

x1 = 1/2

x2 = (-1 - 3) / 4

x2 = -1

The expression is negative over the interval [-1 , 1/2] and is positive for x< -1 and x> 1/2.

We’ve answered 318,911 questions. We can answer yours, too.

Ask a question