If `log_(4) x + log_(8)x^(2) + log_(16)x^(3) = (23)/(2)`, then `log_(x) 8 =`

Options

A. 2
B. `(1)/(2)`
C. 3
D. `(3)/(4)`

Explanation

To solve the equation \( \log_{4} x + \log_{8} x^{2} + \log_{16} x^{3} = \frac{23}{2} \), we will follow these steps:

### Step 1: Rewrite the logarithms in terms of base 2
We know that:
- \( 4 = 2^2 \)
- \( 8 = 2^3 \)
- \( 16 = 2^4 \)

Using the change of base formula, we can rewrite the logarithms:
\[ \log_{4} x = \frac{\log_{2} x}{\log_{2} 4} = \frac{\log_{2} x}{2} \]
\[ \log_{8} x^{2} = \frac{\log_{2} x^{2}}{\log_{2} 8} = \frac{2 \log_{2} x}{3} \]
\[ \log_{16} x^{3} = \frac{\log_{2} x^{3}}{\log_{2} 16} = \frac{3 \log_{2} x}{4} \]

### Step 2: Substitute back into the equation
Now substituting these back into the original equation:
\[ \frac{\log_{2} x}{2} + \frac{2 \log_{2} x}{3} + \frac{3 \log_{2} x}{4} = \frac{23}{2} \]

### Step 3: Find a common denominator
The common denominator for \(2\), \(3\), and \(4\) is \(12\). We will convert each term:
\[ \frac{\log_{2} x}{2} = \frac{6 \log_{2} x}{12} \]
\[ \frac{2 \log_{2} x}{3} = \frac{8 \log_{2} x}{12} \]
\[ \frac{3 \log_{2} x}{4} = \frac{9 \log_{2} x}{12} \]

Now we can combine them:
\[ \frac{6 \log_{2} x + 8 \log_{2} x + 9 \log_{2} x}{12} = \frac{23}{2} \]
\[ \frac{23 \log_{2} x}{12} = \frac{23}{2} \]

### Step 4: Solve for \( \log_{2} x \)
Cross-multiplying gives:
\[ 23 \log_{2} x = 23 \cdot 6 \]
\[ \log_{2} x = 6 \]

### Step 5: Solve for \( x \)
Using the definition of logarithms:
\[ x = 2^{6} = 64 \]

### Step 6: Find \( \log_{x} 8 \)
Now we need to find \( \log_{x} 8 \):
\[ \log_{64} 8 = \frac{\log_{2} 8}{\log_{2} 64} \]
Calculating each logarithm:
\[ \log_{2} 8 = 3 \quad (\text{since } 8 = 2^3) \]
\[ \log_{2} 64 = 6 \quad (\text{since } 64 = 2^6) \]
Thus,
\[ \log_{64} 8 = \frac{3}{6} = \frac{1}{2} \]

### Final Answer
\[ \log_{x} 8 = \frac{1}{2} \]