To solve this inequality, we need to find the values of x that make the expression greater than zero.
First, let's simplify the expression: logx/3(3x^2 - 2x + 1)
Since the logarithm is a function that is only defined for positive values, we can first look at the denominator x/3 and see that x must be greater than 0. Therefore, the domain of the function is x > 0.
Now, let's factor the quadratic expression inside the logarithm: 3x^2 - 2x + 1 = (3x - 1)(x - 1)
Our expression now becomes: logx/3((3x - 1)(x - 1))
Next, let's find the critical points by setting the expression inside the logarithm equal to 0 and solving for x: (3x - 1)(x - 1) = 0 3x - 1 = 0 x = 1/3 x - 1 = 0 x = 1
The critical points are x = 1/3 and x = 1.
Now, we can determine the intervals where the expression is greater than zero by testing values in each interval. The intervals are (0, 1/3), (1/3, 1) and (1, ∞).
For x in the interval (0, 1/3), the expression will be negative as both factors are negative.
For x in the interval (1/3, 1), the expression will be positive as (3x - 1) is positive and (x - 1) is negative.
For x in the interval (1, ∞), the expression will be positive, as both factors are positive.
Therefore, the solution to the inequality logx/3(3x^2 - 2x + 1) > 0 is x ∈ (1/3, 1) ∪ (1, ∞).
To solve this inequality, we need to find the values of x that make the expression greater than zero.
First, let's simplify the expression:
logx/3(3x^2 - 2x + 1)
Since the logarithm is a function that is only defined for positive values, we can first look at the denominator x/3 and see that x must be greater than 0. Therefore, the domain of the function is x > 0.
Now, let's factor the quadratic expression inside the logarithm:
3x^2 - 2x + 1 = (3x - 1)(x - 1)
Our expression now becomes:
logx/3((3x - 1)(x - 1))
Next, let's find the critical points by setting the expression inside the logarithm equal to 0 and solving for x:
(3x - 1)(x - 1) = 0
3x - 1 = 0
x = 1/3
x - 1 = 0
x = 1
The critical points are x = 1/3 and x = 1.
Now, we can determine the intervals where the expression is greater than zero by testing values in each interval. The intervals are (0, 1/3), (1/3, 1) and (1, ∞).
For x in the interval (0, 1/3), the expression will be negative as both factors are negative.
For x in the interval (1/3, 1), the expression will be positive as (3x - 1) is positive and (x - 1) is negative.
For x in the interval (1, ∞), the expression will be positive, as both factors are positive.
Therefore, the solution to the inequality logx/3(3x^2 - 2x + 1) > 0 is x ∈ (1/3, 1) ∪ (1, ∞).