Find range of the function `f(x)=log_(2)[3x-[x+[x+[x]]]]`
(where [.] is greatest integer function)

Explanation

To find the range of the function \( f(x) = \log_2 \left[ 3x - \left[ x + \left[ x + \left[ x \right] \right] \right] \right] \right] \), where \([.]\) denotes the greatest integer function, we will break down the problem step by step.

### Step 1: Simplify the Function
The function can be rewritten as:
\[ f(x) = \log_2 \left[ 3x - 3\left[ x \right] \right] \]
This is because \(\left[ x + \left[ x + \left[ x \right] \right] \right] = 3\left[ x \right]\).

### Step 2: Analyze the Argument of the Logarithm
For the logarithm to be defined, the argument must be greater than 0:
\[ 3x - 3\left[ x \right] > 0 \]
This simplifies to:
\[ 3(x - \left[ x \right]) > 0 \]
Since \(x - \left[ x \right]\) is the fractional part of \(x\), denoted as \(\{x\}\), we have:
\[ 3\{x\} > 0 \quad \Rightarrow \quad \{x\} > 0 \]
This means \(x\) cannot be an integer.

### Step 3: Determine the Values of \(x\)
The fractional part \(\{x\}\) lies in the interval \( (0, 1) \). Therefore, \(x\) can be expressed as:
\[ x = n + r \quad \text{where } n \in \mathbb{Z} \text{ and } 0 < r < 1 \]
This means \(x\) can take any value in the intervals \( (n, n+1) \) for integers \(n\).

### Step 4: Find the Range of \(f(x)\)
Now, substituting \(x = n + r\) into the function:
\[ f(n + r) = \log_2 \left[ 3(n + r) - 3n \right] = \log_2 (3r) \]
Since \(r\) varies from \(0\) to \(1\), \(3r\) varies from \(0\) to \(3\). Therefore, the logarithm will take values from:
\[ \log_2(0) \text{ (undefined) to } \log_2(3) \]
However, since \(r\) cannot be \(0\), we consider the limit as \(r\) approaches \(0\), which leads to:
\[ \lim_{r \to 0^+} \log_2(3r) = -\infty \]
Thus, the range of \(f(x)\) is:
\[ (-\infty, \log_2(3)) \]

### Final Answer
The range of the function \( f(x) = \log_2 \left[ 3x - \left[ x + \left[ x + \left[ x \right] \right] \right] \right] \right] \) is:
\[ (-\infty, \log_2(3)) \]