If `x^([(log_(2)x)^(2)-6log_(2)x+11])=64`, then x=

Options

A. `2`
B. `4`
C. `8`
D. `16`

Answer

To solve the equation \( x^{(\log_2 x)^2 - 6 \log_2 x + 11} = 64 \), we can follow these steps:

### Step 1: Rewrite the equation
We know that \( 64 \) can be expressed as \( 2^6 \). Therefore, we can rewrite the equation as:
\[ x^{(\log_2 x)^2 - 6 \log_2 x + 11} = 2^6 \]

### Step 2: Take logarithm on both sides
Taking logarithm base 2 on both sides, we have:
\[ \log_2\left(x^{(\log_2 x)^2 - 6 \log_2 x + 11}\right) = \log_2(2^6) \]
Using the property of logarithms, this simplifies to:
\[ (\log_2 x)^2 - 6 \log_2 x + 11 = 6 \]

### Step 3: Rearrange the equation
Now, we can rearrange the equation:
\[ (\log_2 x)^2 - 6 \log_2 x + 11 - 6 = 0 \]
This simplifies to:
\[ (\log_2 x)^2 - 6 \log_2 x + 5 = 0 \]

### Step 4: Let \( y = \log_2 x \)
Let \( y = \log_2 x \). The equation becomes:
\[ y^2 - 6y + 5 = 0 \]

### Step 5: Factor the quadratic equation
We can factor this quadratic equation:
\[ (y - 1)(y - 5) = 0 \]
Thus, we have two solutions:
\[ y - 1 = 0 \quad \text{or} \quad y - 5 = 0 \]
This gives us:
\[ y = 1 \quad \text{or} \quad y = 5 \]

### Step 6: Solve for \( x \)
Now, substituting back \( y = \log_2 x \):
1. If \( y = 1 \):
\[ \log_2 x = 1 \implies x = 2^1 = 2 \]
2. If \( y = 5 \):
\[ \log_2 x = 5 \implies x = 2^5 = 32 \]

### Final Answer
Thus, the values of \( x \) are:
\[ x = 2 \quad \text{or} \quad x = 32 \]