The domain of definition of the function
`f(x)=log_(3){-log_(4)((6x-4)/(6x+5))}` , is

Options

A. `(2//3,oo)`
B. `(-oo,-5//6) cup (2//3,oo)`
C. `[2//3,oo)`
D. `(-5//6,2//3)`

Answer

To find the domain of the function \( f(x) = \log_{3}(-\log_{4}\left(\frac{6x-4}{6x+5}\right)) \), we need to ensure that the arguments of both logarithmic functions are valid.

### Step 1: Conditions for the logarithmic functions
1. The base of the logarithm must be greater than zero and not equal to one. Here, the bases are 3 and 4, which are both valid.
2. The argument of the logarithm must be greater than zero.

### Step 2: Analyze the inner logarithm
We start with the inner logarithm:
\[ -\log_{4}\left(\frac{6x-4}{6x+5}\right) > 0 \]
This implies:
\[ \log_{4}\left(\frac{6x-4}{6x+5}\right) < 0 \]
For the logarithm to be negative, the argument must be between 0 and 1:
\[ 0 < \frac{6x-4}{6x+5} < 1 \]

### Step 3: Solve the inequality \( \frac{6x-4}{6x+5} > 0 \)
The fraction \( \frac{6x-4}{6x+5} \) is positive when both the numerator and denominator are either both positive or both negative.

**Numerator:**
\[ 6x - 4 > 0 \implies x > \frac{2}{3} \]

**Denominator:**
\[ 6x + 5 > 0 \implies x > -\frac{5}{6} \]

### Step 4: Analyze the intervals
1. **For \( x > \frac{2}{3} \)**: Both the numerator and denominator are positive.
2. **For \( x < -\frac{5}{6} \)**: Both the numerator and denominator are negative.

Thus, the solution for \( \frac{6x-4}{6x+5} > 0 \) is:
\[ x \in (-\infty, -\frac{5}{6}) \cup \left(\frac{2}{3}, \infty\right) \]

### Step 5: Solve the inequality \( \frac{6x-4}{6x+5} < 1 \)
We rearrange the inequality:
\[ \frac{6x-4}{6x+5} - 1 < 0 \implies \frac{6x-4 - (6x+5)}{6x+5} < 0 \]
This simplifies to:
\[ \frac{-9}{6x+5} < 0 \]

The fraction \( \frac{-9}{6x+5} < 0 \) is negative when the denominator is positive:
\[ 6x + 5 > 0 \implies x > -\frac{5}{6} \]

### Step 6: Combine the results
Now we combine the results from both inequalities:
1. From \( \frac{6x-4}{6x+5} > 0 \): \( x \in (-\infty, -\frac{5}{6}) \cup \left(\frac{2}{3}, \infty\right) \)
2. From \( \frac{6x-4}{6x+5} < 1 \): \( x > -\frac{5}{6} \)

The intersection of these two sets gives us:
\[ x \in \left(\frac{2}{3}, \infty\right) \]

### Final Answer
Thus, the domain of the function \( f(x) \) is:
\[ \boxed{\left(\frac{2}{3}, \infty\right)} \]