If `x_(1),x_(2) "are two solutions of" X^(lnx^(2) )=e^(18) "then product of" X_(1)X_(2)` can be equal to

Options

A. `(sec^(2)theta+co s ec^(2)theta)/(sec^(2)theta.c os ec^(2)theta)`,(wherever defined)
B. `(cot^(2)18^(@)-cvos^(2)18^(@))/(cost^(2)18^(@)cot^(2)18^(@))`
C. `sec 0+sec((pi)/(5))+sec((2pi)/(5))+sec((3pi)/(5))+sec((4pi)/(5))`
D. `2log6^(3).3^(log6^(3))^(-1)`

Answer

To solve the equation \( x^{\ln x^2} = e^{18} \) and find the product of its solutions \( x_1 \) and \( x_2 \), we will follow these steps:

### Step 1: Simplify the equation
The equation can be rewritten using properties of logarithms. We know that \( \ln x^2 = 2 \ln x \). Thus, we can rewrite the equation as:
\[ x^{2 \ln x} = e^{18} \]

### Step 2: Take the natural logarithm of both sides
Taking the natural logarithm of both sides gives us:
\[ \ln(x^{2 \ln x}) = \ln(e^{18}) \]
Using the property of logarithms \( \ln(a^b) = b \ln a \), we can simplify the left side:
\[ 2 \ln x \cdot \ln x = 18 \]
This simplifies to:
\[ 2 (\ln x)^2 = 18 \]

### Step 3: Solve for \( \ln x \)
Dividing both sides by 2:
\[ (\ln x)^2 = 9 \]
Taking the square root of both sides gives us:
\[ \ln x = 3 \quad \text{or} \quad \ln x = -3 \]

### Step 4: Exponentiate to solve for \( x \)
Exponentiating both sides, we find:
1. From \( \ln x = 3 \):
\[ x_1 = e^3 \]
2. From \( \ln x = -3 \):
\[ x_2 = e^{-3} \]

### Step 5: Calculate the product \( x_1 x_2 \)
Now, we calculate the product of the two solutions:
\[ x_1 x_2 = e^3 \cdot e^{-3} = e^{3 - 3} = e^0 = 1 \]

### Final Answer
Thus, the product of the solutions \( x_1 x_2 \) is:
\[ \boxed{1} \]