If `x=log_2(sqrt(56+sqrt(56+sqrt(56+sqrt(56+ . . . oo)))))` then which of the following statement holds good?

Options

A. `x lt 0`
B. `0 lt x lt 2`
C. `2 lt x lt 4`
D. `3 lt x lt 4`

Explanation

To solve the problem, we start with the expression:

\[ x = \log_2\left(\sqrt{56 + \sqrt{56 + \sqrt{56 + \sqrt{56 + \ldots}}}}\right) \]

Let's denote the infinite nested square root as \( y \):

\[ y = \sqrt{56 + \sqrt{56 + \sqrt{56 + \ldots}}} \]

### Step 1: Set up the equation for \( y \)

From the definition of \( y \), we can write:

\[ y = \sqrt{56 + y} \]

### Step 2: Square both sides

Now, squaring both sides to eliminate the square root gives us:

\[ y^2 = 56 + y \]

### Step 3: Rearrange the equation

Rearranging this equation leads to:

\[ y^2 - y - 56 = 0 \]

### Step 4: Solve the quadratic equation

Now we will solve the quadratic equation \( y^2 - y - 56 = 0 \) using the factoring method. We need to find two numbers that multiply to \(-56\) and add up to \(-1\). The numbers are \( -8 \) and \( 7 \). Thus, we can factor the equation as:

\[ (y - 8)(y + 7) = 0 \]

### Step 5: Find the roots

Setting each factor to zero gives us:

1. \( y - 8 = 0 \) → \( y = 8 \)
2. \( y + 7 = 0 \) → \( y = -7 \) (not valid since \( y \) must be positive)

Thus, we have:

\[ y = 8 \]

### Step 6: Substitute back to find \( x \)

Now substituting \( y \) back into the expression for \( x \):

\[ x = \log_2(8) \]

### Step 7: Simplify \( x \)

Since \( 8 = 2^3 \), we can simplify:

\[ x = \log_2(2^3) = 3 \]

### Conclusion

Thus, we find that:

\[ x = 3 \]

### Step 8: Analyze the options

Now we need to check the statements regarding \( x \):

1. \( x < 0 \) → False (since \( x = 3 \))
2. \( 0 < x < 2 \) → False (since \( x = 3 \))
3. \( 2 < x < 4 \) → True (since \( x = 3 \))
4. \( 3 < x < 4 \) → False (since \( x = 3 \))

The correct statement is that \( x \) is greater than 2 and less than 4.