If `x = sqrt(7)+(1)/(sqrt(7))`, then the value of `(128)^(x^(2))` is-

Options

A. `2^(40)`
B. `2^(64)`
C. `2^(56)`
D. `2^(24)`

Answer

To solve the problem, we need to find the value of \( (128)^{x^2} \) where \( x = \sqrt{7} + \frac{1}{\sqrt{7}} \).

### Step 1: Calculate \( x^2 \)
We start by squaring \( x \):
\[ x^2 = \left( \sqrt{7} + \frac{1}{\sqrt{7}} \right)^2 \]
Using the identity \( (a + b)^2 = a^2 + 2ab + b^2 \), we can expand this:
\[ x^2 = \left( \sqrt{7} \right)^2 + 2 \left( \sqrt{7} \cdot \frac{1}{\sqrt{7}} \right) + \left( \frac{1}{\sqrt{7}} \right)^2 \]
Calculating each term:
- \( \left( \sqrt{7} \right)^2 = 7 \)
- \( 2 \left( \sqrt{7} \cdot \frac{1}{\sqrt{7}} \right) = 2 \)
- \( \left( \frac{1}{\sqrt{7}} \right)^2 = \frac{1}{7} \)

Putting it all together:
\[ x^2 = 7 + 2 + \frac{1}{7} = 9 + \frac{1}{7} \]
To combine these, we convert 9 into a fraction:
\[ 9 = \frac{63}{7} \]
Thus:
\[ x^2 = \frac{63}{7} + \frac{1}{7} = \frac{64}{7} \]

### Step 2: Substitute \( x^2 \) into \( (128)^{x^2} \)
Now we need to find \( (128)^{x^2} \):
\[ (128)^{x^2} = (128)^{\frac{64}{7}} \]
We can express 128 as a power of 2:
\[ 128 = 2^7 \]
So we can rewrite the expression:
\[ (128)^{\frac{64}{7}} = (2^7)^{\frac{64}{7}} \]
Using the power of a power property \( (a^m)^n = a^{m \cdot n} \):
\[ (2^7)^{\frac{64}{7}} = 2^{7 \cdot \frac{64}{7}} = 2^{64} \]

### Final Answer
Thus, the value of \( (128)^{x^2} \) is:
\[ \boxed{2^{64}} \]