Solve the equation `(2+sqrt(3))^(x^(2)-2x+1)+(2-sqrt(3))^(x^(2)-2x-1)=101/(10(2-sqrt(3))`

Explanation

To solve the equation

\[ (2+\sqrt{3})^{x^2-2x+1} + (2-\sqrt{3})^{x^2-2x-1} = \frac{101}{10(2-\sqrt{3})} \]

we will follow these steps:

### Step 1: Rewrite the equation
First, we rewrite the equation in a more manageable form:

\[ (2+\sqrt{3})^{(x-1)^2} + (2-\sqrt{3})^{(x^2-2x-1)} = \frac{101}{10(2-\sqrt{3})} \]

### Step 2: Simplify the right-hand side
We can simplify the right-hand side by recognizing that \(2 - \sqrt{3} = \frac{1}{2+\sqrt{3}}\). Thus, we can rewrite the right-hand side as:

\[ \frac{101}{10} \cdot \frac{1}{2+\sqrt{3}} = \frac{101}{10(2+\sqrt{3})} \]

### Step 3: Substitute values
Let \( t = (2+\sqrt{3})^{(x-1)^2} \). Then, we have:

\[ t + \frac{1}{t} = \frac{101}{10(2+\sqrt{3})} \]

### Step 4: Multiply through by \(t\)
Multiplying through by \(t\) gives:

\[ t^2 + 1 = \frac{101}{10(2+\sqrt{3})} t \]

### Step 5: Rearrange into standard form
Rearranging gives us a quadratic equation:

\[ t^2 - \frac{101}{10(2+\sqrt{3})} t + 1 = 0 \]

### Step 6: Use the quadratic formula
Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -\frac{101}{10(2+\sqrt{3})}\), and \(c = 1\):

\[ t = \frac{\frac{101}{10(2+\sqrt{3})} \pm \sqrt{\left(\frac{101}{10(2+\sqrt{3})}\right)^2 - 4}}{2} \]

### Step 7: Solve for \(t\)
Calculate the discriminant:

\[ \left(\frac{101}{10(2+\sqrt{3})}\right)^2 - 4 \]

Then substitute back to find \(t\).

### Step 8: Solve for \(x\)
Once we find \(t\), we can revert back to find \(x\) using:

\[ (2+\sqrt{3})^{(x-1)^2} = t \]

Taking logarithms on both sides gives:

\[ (x-1)^2 = \log_{(2+\sqrt{3})}(t) \]

Finally, solving for \(x\):

\[ x - 1 = \pm \sqrt{\log_{(2+\sqrt{3})}(t)} \]
\[ x = 1 \pm \sqrt{\log_{(2+\sqrt{3})}(t)} \]

### Final Answer
Thus, the solutions for \(x\) are:

\[ x_1 = 1 + \sqrt{\log_{(2+\sqrt{3})}(t)}, \quad x_2 = 1 - \sqrt{\log_{(2+\sqrt{3})}(t)} \]