The maximum integral values of x in the domain of `f (x) =log _(10)(log _(1//3)(log _(4) (x-5))` is : (a). 5 (b). 7 (c). 8 (d). 9

Options

A. 5
B. 7
C. 8
D. 9

Answer

To find the maximum integral values of \( x \) in the domain of the function \( f(x) = \log_{10}(\log_{\frac{1}{3}}(\log_{4}(x-5))) \), we need to ensure that the arguments of all logarithmic functions are positive.

### Step 1: Determine the innermost logarithm
We start with the innermost logarithm, which is \( \log_{4}(x-5) \). For this logarithm to be defined, the argument must be positive:
\[ x - 5 > 0 \implies x > 5 \]

**Hint:** Ensure the argument of the innermost logarithm is greater than zero.

### Step 2: Set conditions for the second logarithm
Next, we consider the logarithm \( \log_{\frac{1}{3}}(\log_{4}(x-5)) \). For this logarithm to be defined, its argument must also be positive:
\[ \log_{4}(x-5) > 0 \]
This implies:
\[ x - 5 > 4^0 \implies x - 5 > 1 \implies x > 6 \]

**Hint:** Remember that \( \log_{b}(y) > 0 \) means \( y > 1 \) when \( 0 < b < 1 \).

### Step 3: Set conditions for the outermost logarithm
Now, we look at the outermost logarithm \( \log_{10}(\log_{\frac{1}{3}}(\log_{4}(x-5))) \). For this logarithm to be defined, its argument must also be positive:
\[ \log_{\frac{1}{3}}(\log_{4}(x-5)) > 0 \]
This means:
\[ \log_{4}(x-5) < 1 \]
This leads to:
\[ x - 5 < 4^1 \implies x - 5 < 4 \implies x < 9 \]

**Hint:** Ensure that the argument of the logarithm is less than 1 when the base is between 0 and 1.

### Step 4: Combine the inequalities
From the above steps, we have two inequalities:
1. \( x > 6 \)
2. \( x < 9 \)

Combining these gives:
\[ 6 < x < 9 \]

### Step 5: Identify the maximum integral value
The integral values of \( x \) that satisfy \( 6 < x < 9 \) are \( 7 \) and \( 8 \). The maximum integral value is therefore:
\[ \text{Maximum integral value of } x = 8 \]

### Conclusion
Thus, the maximum integral value of \( x \) in the domain of \( f(x) \) is \( 8 \).

**Final Answer:** (c) 8

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