If `log_(2) 6 + (1)/(2x ) =log_(2) (2^(1//x) + 8)`, then the values of x are

Options

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

Answer

To solve the equation \( \log_{2} 6 + \frac{1}{2x} = \log_{2} (2^{\frac{1}{x}} + 8) \), we will follow these steps:

### Step 1: Isolate the logarithmic terms
We can start by isolating the logarithmic term on one side of the equation. This gives us:
\[ \frac{1}{2x} = \log_{2} (2^{\frac{1}{x}} + 8) - \log_{2} 6 \]

### Step 2: Use the properties of logarithms
Using the property of logarithms that states \( \log_{a} b - \log_{a} c = \log_{a} \left( \frac{b}{c} \right) \), we can rewrite the right side:
\[ \frac{1}{2x} = \log_{2} \left( \frac{2^{\frac{1}{x}} + 8}{6} \right) \]

### Step 3: Exponentiate both sides
Next, we exponentiate both sides to eliminate the logarithm:
\[ 2^{\frac{1}{2x}} = \frac{2^{\frac{1}{x}} + 8}{6} \]

### Step 4: Multiply both sides by 6
To simplify, we multiply both sides by 6:
\[ 6 \cdot 2^{\frac{1}{2x}} = 2^{\frac{1}{x}} + 8 \]

### Step 5: Rearrange the equation
Rearranging gives us:
\[ 2^{\frac{1}{x}} = 6 \cdot 2^{\frac{1}{2x}} - 8 \]

### Step 6: Substitute \( t = 2^{\frac{1}{x}} \)
Let \( t = 2^{\frac{1}{x}} \). Then \( 2^{\frac{1}{2x}} = \sqrt{t} \):
\[ t = 6\sqrt{t} - 8 \]

### Step 7: Rearrange into a quadratic equation
Rearranging gives us:
\[ t^2 - 6t + 8 = 0 \]

### Step 8: Factor the quadratic equation
Factoring the quadratic:
\[ (t - 2)(t - 4) = 0 \]

### Step 9: Solve for \( t \)
Setting each factor to zero gives:
\[ t - 2 = 0 \quad \Rightarrow \quad t = 2 \]
\[ t - 4 = 0 \quad \Rightarrow \quad t = 4 \]

### Step 10: Back-substitute for \( x \)
Now we back-substitute for \( x \):
1. For \( t = 2 \):
\[ 2^{\frac{1}{x}} = 2 \quad \Rightarrow \quad \frac{1}{x} = 1 \quad \Rightarrow \quad x = 1 \]

2. For \( t = 4 \):
\[ 2^{\frac{1}{x}} = 4 \quad \Rightarrow \quad \frac{1}{x} = 2 \quad \Rightarrow \quad x = \frac{1}{2} \]

### Final Answer
The values of \( x \) are \( \frac{1}{2} \) and \( 1 \).