Let `S equiv {x in R : (sqrt 3 + sqrt 2)^x + (sqrt 3 - sqrt 2)^x = 10}`. Then the number of elements in S is :

Options

A. 4
B. 0
C. 2
D. 1

Answer

To solve the equation \((\sqrt{3} + \sqrt{2})^x + (\sqrt{3} - \sqrt{2})^x = 10\), we can follow these steps:

### Step 1: Let \( t = (\sqrt{3} + \sqrt{2})^x \)

By substituting \( t \), we can express \((\sqrt{3} - \sqrt{2})^x\) in terms of \( t \):
\[ (\sqrt{3} - \sqrt{2})^x = \left(\frac{1}{\sqrt{3} + \sqrt{2}}\right)^x = \frac{1}{t} \]

### Step 2: Rewrite the equation

Now, substituting back into the original equation, we have:
\[ t + \frac{1}{t} = 10 \]

### Step 3: Multiply through by \( t \)

To eliminate the fraction, multiply both sides by \( t \):
\[ t^2 + 1 = 10t \]

### Step 4: Rearrange into standard quadratic form

Rearranging gives us:
\[ t^2 - 10t + 1 = 0 \]

### Step 5: Solve the quadratic equation

Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1, b = -10, c = 1 \):
\[ t = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{10 \pm \sqrt{100 - 4}}{2} = \frac{10 \pm \sqrt{96}}{2} \]
\[ t = \frac{10 \pm 4\sqrt{6}}{2} = 5 \pm 2\sqrt{6} \]

### Step 6: Find the values of \( x \)

Now we have two cases for \( t \):
1. \( t = 5 + 2\sqrt{6} \)
2. \( t = 5 - 2\sqrt{6} \)

Since \( t = (\sqrt{3} + \sqrt{2})^x \), we take the logarithm to find \( x \):
1. For \( t = 5 + 2\sqrt{6} \):
\[ x = \log_{\sqrt{3} + \sqrt{2}}(5 + 2\sqrt{6}) \]
2. For \( t = 5 - 2\sqrt{6} \):
\[ x = \log_{\sqrt{3} + \sqrt{2}}(5 - 2\sqrt{6}) \]

### Step 7: Determine the number of solutions

Since both \( 5 + 2\sqrt{6} > 0 \) and \( 5 - 2\sqrt{6} > 0 \) (as \( 2\sqrt{6} \) is less than 5), both values of \( t \) yield valid real solutions for \( x \).

Thus, the number of elements in the set \( S \) is **2**.

### Final Answer:
The number of elements in \( S \) is **2**.

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