To solve this inequality, we first need to rewrite it in a more simplified form. The given inequality is:
lg^2(x) - lg(x) - 6 ≥ 0
Let's substitute lg(x) as t:
t = lg(x)
Now, the inequality becomes:
t^2 - t - 6 ≥ 0
We can rewrite this quadratic inequality in factored form:
(t - 3)(t + 2) ≥ 0
Now, we need to find the critical points by setting each factor to zero:
t - 3 = 0t = 3
t + 2 = 0t = -2
So, the critical points are t = 3 and t = -2. We can now create intervals on the number line to test the sign of the expression:
-∞ -2 3 +∞
From the number line, we can see that the inequality is true for t ≤ -2 and t ≥ 3. Now, let's substitute back t as lg(x):
lg(x) ≤ -2 and lg(x) ≥ 3
Solving for x:
x ≤ 10^(-2) and x ≥ 10^3
x ≤ 0.01 and x ≥ 1000
Therefore, the solution for the inequality lg^2(x) - lg(x) - 6 ≥ 0 is:
x ≤ 0.01 or x ≥ 1000.
To solve this inequality, we first need to rewrite it in a more simplified form. The given inequality is:
lg^2(x) - lg(x) - 6 ≥ 0
Let's substitute lg(x) as t:
t = lg(x)
Now, the inequality becomes:
t^2 - t - 6 ≥ 0
We can rewrite this quadratic inequality in factored form:
(t - 3)(t + 2) ≥ 0
Now, we need to find the critical points by setting each factor to zero:
t - 3 = 0
t = 3
t + 2 = 0
t = -2
So, the critical points are t = 3 and t = -2. We can now create intervals on the number line to test the sign of the expression:
-∞ -2 3 +∞
| + |From the number line, we can see that the inequality is true for t ≤ -2 and t ≥ 3. Now, let's substitute back t as lg(x):
lg(x) ≤ -2 and lg(x) ≥ 3
Solving for x:
x ≤ 10^(-2) and x ≥ 10^3
x ≤ 0.01 and x ≥ 1000
Therefore, the solution for the inequality lg^2(x) - lg(x) - 6 ≥ 0 is:
x ≤ 0.01 or x ≥ 1000.