The solution set of the equation `8x= e^(x^(2)+log(-x))` is

Options

A. `phi`
B. `x in R`
C. `x in I`
D. `x in (-oo, 0)`

Explanation

To solve the equation \( 8x = e^{x^2 + \log(-x)} \), we will analyze both sides of the equation step by step.

### Step 1: Determine the domain of the logarithm
The term \( \log(-x) \) requires that the argument \(-x\) be positive. Therefore, we need:
\[ -x > 0 \implies x < 0 \]
This means that \( x \) must be a negative number.

**Hint:** Remember that the logarithm is only defined for positive arguments.

### Step 2: Analyze the left-hand side (LHS)
The left-hand side of the equation is \( 8x \). Since we established that \( x < 0 \), we can conclude:
\[ 8x < 0 \]
This means that the left-hand side is negative.

**Hint:** Multiplying a negative number by a positive constant results in a negative number.

### Step 3: Analyze the right-hand side (RHS)
The right-hand side of the equation is \( e^{x^2 + \log(-x)} \). The exponential function \( e^y \) is always positive for any real number \( y \). Therefore:
\[ e^{x^2 + \log(-x)} > 0 \]
This means that the right-hand side is positive.

**Hint:** The exponential function is always greater than zero, regardless of the exponent.

### Step 4: Compare LHS and RHS
Now we compare the two sides:
- LHS: \( 8x < 0 \) (negative)
- RHS: \( e^{x^2 + \log(-x)} > 0 \) (positive)

Since the left-hand side is negative and the right-hand side is positive, there cannot be any value of \( x \) that satisfies the equation.

**Hint:** A negative number can never equal a positive number.

### Conclusion
Since there are no values of \( x \) that satisfy the equation, the solution set is the null set.

**Final Answer:** The solution set is \( \emptyset \) (null set).