Question

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

• A. 4
• B. 0
• C. 2
• D. 1

Answer 1

The Correct Option is C

Consider:

\[ \left( \sqrt{3} + \sqrt{2} \right)^x + \left( \sqrt{3} - \sqrt{2} \right)^x = 10 \]

Let:

\[ \left( \sqrt{3} + \sqrt{2} \right)^x = t \]

Thus:

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

Substitute and simplify:

\[ t + \frac{1}{t} = 10 \]

Multiplying through by \( t \) gives:

\[ t^2 - 10t + 1 = 0 \]

Solving this quadratic equation:

\[ t = \frac{10 \pm \sqrt{100 - 4}}{2} = \frac{10 \pm \sqrt{96}}{2} = \frac{10 \pm 4\sqrt{6}}{2} = 5 \pm 2\sqrt{6} \]

Since:

\[ \left( \sqrt{3} + \sqrt{2} \right)^x > 0, \quad t = 5 + 2\sqrt{6} \]

Thus, the corresponding values of \( x \) are:
\[ x = 2 \quad \text{or} \quad x = -2 \]

Number of solutions = 2.

Answer 2

Determine the number of real solutions \( x \) to the equation \( (\sqrt{3} + \sqrt{2})^x + (\sqrt{3} - \sqrt{2})^x = 10 \).

Concept Used:

Let \( t = (\sqrt{3} + \sqrt{2})^x \). Then \( (\sqrt{3} - \sqrt{2})^x = \frac{1}{t} \) because \( (\sqrt{3} - \sqrt{2}) = \frac{1}{\sqrt{3} + \sqrt{2}} \). The equation becomes \( t + \frac{1}{t} = 10 \), which simplifies to \( t^2 - 10t + 1 = 0 \).

Step-by-Step Solution:

Step 1: Make the substitution \( t = (\sqrt{3} + \sqrt{2})^x \).

Since \( (\sqrt{3} - \sqrt{2}) = \frac{1}{\sqrt{3} + \sqrt{2}} \), we have \( (\sqrt{3} - \sqrt{2})^x = \frac{1}{t} \).

The equation becomes:

\[ t + \frac{1}{t} = 10 \]

Step 2: Solve for \( t \).

\[ t + \frac{1}{t} = 10 \implies t^2 - 10t + 1 = 0 \] \[ t = \frac{10 \pm \sqrt{100 - 4}}{2} = \frac{10 \pm \sqrt{96}}{2} = \frac{10 \pm 4\sqrt{6}}{2} = 5 \pm 2\sqrt{6} \]

Step 3: Check the values of \( t \).

Note: \( \sqrt{3} + \sqrt{2} > 1 \), so \( t = (\sqrt{3} + \sqrt{2})^x > 0 \) for all real \( x \).

Both \( 5 + 2\sqrt{6} \) and \( 5 - 2\sqrt{6} \) are positive.

Also, \( (5 + 2\sqrt{6})(5 - 2\sqrt{6}) = 25 - 24 = 1 \), so \( 5 - 2\sqrt{6} = \frac{1}{5 + 2\sqrt{6}} \).

Step 4: Solve for \( x \) in each case.

Case 1: \( (\sqrt{3} + \sqrt{2})^x = 5 + 2\sqrt{6} \)

Note: \( 5 + 2\sqrt{6} = (\sqrt{3} + \sqrt{2})^2 \) because \( (\sqrt{3} + \sqrt{2})^2 = 3 + 2 + 2\sqrt{6} = 5 + 2\sqrt{6} \).

So \( (\sqrt{3} + \sqrt{2})^x = (\sqrt{3} + \sqrt{2})^2 \implies x = 2 \).

Case 2: \( (\sqrt{3} + \sqrt{2})^x = 5 - 2\sqrt{6} \)

But \( 5 - 2\sqrt{6} = (\sqrt{3} - \sqrt{2})^2 \) because \( (\sqrt{3} - \sqrt{2})^2 = 3 + 2 - 2\sqrt{6} = 5 - 2\sqrt{6} \).

Also, \( (\sqrt{3} - \sqrt{2})^2 = \frac{1}{(\sqrt{3} + \sqrt{2})^2} \).

So \( (\sqrt{3} + \sqrt{2})^x = (\sqrt{3} + \sqrt{2})^{-2} \implies x = -2 \).

Step 5: Count the number of solutions.

We found \( x = 2 \) and \( x = -2 \). Both are real numbers.

Thus, the set \( S \) has 2 elements.

Therefore, the number of elements in \( S \) is 2.