The value of a for which the equation `(sqrt(a+sqrt(a^(2)-1)))^(x)+(sqrt(a-sqrt(a^(2)-1)))^(x)=2a` has only two solutions is …...

Answer

To find the value of \( a \) for which the equation

\[ \left(\sqrt{a + \sqrt{a^2 - 1}}\right)^x + \left(\sqrt{a - \sqrt{a^2 - 1}}\right)^x = 2a \]

has only two solutions, we can follow these steps:

### Step 1: Define the Terms
Let
\[ y = \sqrt{a + \sqrt{a^2 - 1}} \quad \text{and} \quad z = \sqrt{a - \sqrt{a^2 - 1}}. \]
Then the equation can be rewritten as:
\[ y^x + z^x = 2a. \]

### Step 2: Relate \( y \) and \( z \)
Notice that:
\[ y \cdot z = \sqrt{(a + \sqrt{a^2 - 1})(a - \sqrt{a^2 - 1})} = \sqrt{a^2 - (a^2 - 1)} = \sqrt{1} = 1. \]
This implies that \( z = \frac{1}{y} \).

### Step 3: Substitute \( z \) in the Equation
Substituting \( z \) into the equation gives:
\[ y^x + \left(\frac{1}{y}\right)^x = 2a. \]
This simplifies to:
\[ y^x + \frac{1}{y^x} = 2a. \]

### Step 4: Let \( t = y^x \)
Let \( t = y^x \). Then the equation becomes:
\[ t + \frac{1}{t} = 2a. \]
Multiplying through by \( t \) gives:
\[ t^2 - 2at + 1 = 0. \]

### Step 5: Analyze the Quadratic Equation
For the quadratic equation \( t^2 - 2at + 1 = 0 \) to have exactly two solutions, the discriminant must be zero. The discriminant \( D \) is given by:
\[ D = (2a)^2 - 4 \cdot 1 \cdot 1 = 4a^2 - 4. \]
Setting the discriminant to zero for the quadratic to have one double root:
\[ 4a^2 - 4 = 0. \]
This simplifies to:
\[ 4a^2 = 4 \implies a^2 = 1 \implies a = 1 \text{ or } a = -1. \]

### Step 6: Determine Valid Values for \( a \)
Since \( a \) must be positive (as it is under a square root), we take:
\[ a = 1. \]

### Conclusion
Thus, the value of \( a \) for which the equation has only two solutions is:
\[ \boxed{1}. \]