The number of real solution of equation `Sin(e^x) = 5^x + 5^(-x)` is :

Options

A. 0
B. 1
C. 2
D. Infinitely many

Answer

To find the number of real solutions for the equation \( \sin(e^x) = 5^x + 5^{-x} \), we will analyze both sides of the equation step by step.

### Step 1: Analyze the Left-Hand Side (LHS)
The left-hand side of the equation is \( \sin(e^x) \).

- The sine function, \( \sin(x) \), has a range of \([-1, 1]\) for all real numbers \(x\).
- Since \(e^x\) is always positive (greater than 0), \( \sin(e^x) \) will also be bounded within \([-1, 1]\).

**Conclusion for LHS**:
\[ -1 \leq \sin(e^x) \leq 1 \]

**Hint**: Remember that the sine function oscillates between -1 and 1 regardless of the input.

### Step 2: Analyze the Right-Hand Side (RHS)
The right-hand side of the equation is \( 5^x + 5^{-x} \).

- We can rewrite \( 5^{-x} \) as \( \frac{1}{5^x} \).
- Thus, we have:
\[ 5^x + 5^{-x} = 5^x + \frac{1}{5^x} \]

**Step 2.1: Use the AM-GM Inequality**
Applying the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
\[ \frac{5^x + 5^{-x}}{2} \geq \sqrt{5^x \cdot 5^{-x}} = \sqrt{1} = 1 \]
This implies:
\[ 5^x + 5^{-x} \geq 2 \]

**Conclusion for RHS**:
\[ 5^x + 5^{-x} \geq 2 \]

**Hint**: The AM-GM inequality states that the arithmetic mean of non-negative numbers is greater than or equal to their geometric mean.

### Step 3: Compare LHS and RHS
Now we compare the conclusions from both sides:

- From the LHS, we have:
\[ -1 \leq \sin(e^x) \leq 1 \]
- From the RHS, we have:
\[ 5^x + 5^{-x} \geq 2 \]

### Step 4: Conclusion
Since the maximum value of the left-hand side is 1 and the minimum value of the right-hand side is 2, we can conclude that:
\[ \sin(e^x) \leq 1 < 2 \leq 5^x + 5^{-x} \]

This means there is no value of \(x\) for which the equation \( \sin(e^x) = 5^x + 5^{-x} \) holds true.

**Final Answer**: The number of real solutions is **0**.