If the fourth term in the binomial expansin of `((2)/(x) + x^(log_(8) x))^(6) (x gt 0)` is `20 xx 8^(7)`, then the value of x is

Options

A. `8^(-2)`
B. 8
C. `8^(2)`
D. `8^(3)`

Explanation

To solve the problem, we need to find the value of \( x \) given that the fourth term in the binomial expansion of \( \left( \frac{2}{x} + x^{\log_{8} x} \right)^{6} \) is equal to \( 20 \times 8^{7} \).

### Step-by-Step Solution:

1. **Identify the General Term in the Binomial Expansion**:
The general term \( T_{r+1} \) in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, \( a = \frac{2}{x} \), \( b = x^{\log_{8} x} \), and \( n = 6 \).

2. **Find the Fourth Term**:
The fourth term corresponds to \( r = 3 \) (since \( T_{r+1} \) corresponds to \( r \)):
\[ T_{4} = \binom{6}{3} \left(\frac{2}{x}\right)^{6-3} \left(x^{\log_{8} x}\right)^{3} \]
Simplifying this gives:
\[ T_{4} = \binom{6}{3} \left(\frac{2}{x}\right)^{3} \left(x^{\log_{8} x}\right)^{3} \]

3. **Calculate the Binomial Coefficient**:
The binomial coefficient \( \binom{6}{3} \) is calculated as:
\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]
Thus,
\[ T_{4} = 20 \left(\frac{2}{x}\right)^{3} \left(x^{\log_{8} x}\right)^{3} \]

4. **Simplify the Expression**:
This can be rewritten as:
\[ T_{4} = 20 \cdot \frac{2^3}{x^3} \cdot x^{3 \log_{8} x} = 20 \cdot \frac{8}{x^3} \cdot x^{3 \log_{8} x} \]
Which simplifies to:
\[ T_{4} = 20 \cdot 8 \cdot x^{3 \log_{8} x - 3} \]

5. **Set the Expression Equal to Given Value**:
We know that:
\[ T_{4} = 20 \times 8^{7} \]
Therefore, we can set up the equation:
\[ 20 \cdot 8 \cdot x^{3 \log_{8} x - 3} = 20 \cdot 8^{7} \]

6. **Cancel Common Terms**:
Dividing both sides by \( 20 \):
\[ 8 \cdot x^{3 \log_{8} x - 3} = 8^{7} \]
Dividing both sides by \( 8 \):
\[ x^{3 \log_{8} x - 3} = 8^{6} \]

7. **Rewrite \( 8^{6} \)**:
Since \( 8 = 2^3 \), we have:
\[ 8^{6} = (2^3)^{6} = 2^{18} \]

8. **Express \( x^{3 \log_{8} x - 3} \)**:
We can express \( 8^{6} \) in terms of base \( 2 \):
\[ x^{3 \log_{8} x - 3} = 2^{18} \]

9. **Taking Logarithms**:
Taking logarithm base \( 2 \) on both sides gives:
\[ (3 \log_{8} x - 3) \log_{2} x = 18 \]

10. **Substituting \( \log_{8} x \)**:
Recall that \( \log_{8} x = \frac{\log_{2} x}{\log_{2} 8} = \frac{\log_{2} x}{3} \):
\[ 3 \cdot \frac{\log_{2} x}{3} \log_{2} x - 3 \log_{2} x = 18 \]
Simplifying gives:
\[ (\log_{2} x)^2 - 3 \log_{2} x - 18 = 0 \]

11. **Solve the Quadratic Equation**:
Let \( y = \log_{2} x \):
\[ y^2 - 3y - 18 = 0 \]
Using the quadratic formula:
\[ y = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-18)}}{2 \cdot 1} = \frac{3 \pm \sqrt{81}}{2} = \frac{3 \pm 9}{2} \]
This gives:
\[ y = 6 \quad \text{or} \quad y = -3 \]

12. **Find \( x \)**:
- If \( y = 6 \): \( \log_{2} x = 6 \) implies \( x = 2^6 = 64 \).
- If \( y = -3 \): \( \log_{2} x = -3 \) implies \( x = 2^{-3} = \frac{1}{8} \).

Since \( x > 0 \), both values are valid, but based on the context of the problem, we take \( x = 64 \).

### Final Answer:
The value of \( x \) is \( 64 \).