The equation `2^(|x^(2)-12|)=sqrt(e^(|x|log4))` has

Options

A. no real solution
B. only two real solutions whose sum is zero
C. only two real solutions whose sum is not zero
D. four real solutions whose sum is zero.

Explanation

To solve the equation \( 2^{|x^2 - 12|} = \sqrt{e^{|x| \log 4}} \), we will follow a step-by-step approach.

### Step 1: Simplify the Right-Hand Side
The right-hand side can be simplified. Recall that \( \sqrt{a} = a^{1/2} \). Therefore,
\[ \sqrt{e^{|x| \log 4}} = e^{\frac{1}{2} |x| \log 4} \]
Using the property \( e^{\log a} = a \), we can rewrite \( e^{\frac{1}{2} |x| \log 4} \) as:
\[ e^{\frac{1}{2} |x| \log 4} = 4^{\frac{1}{2} |x|} = 2^{|x|} \]
Thus, the equation becomes:
\[ 2^{|x^2 - 12|} = 2^{|x|} \]

### Step 2: Set the Exponents Equal
Since the bases are the same, we can set the exponents equal to each other:
\[ |x^2 - 12| = |x| \]

### Step 3: Consider Cases for Absolute Values
We need to consider two cases for the absolute values.

#### Case 1: \( x^2 - 12 = x \)
Rearranging gives:
\[ x^2 - x - 12 = 0 \]
Factoring the quadratic:
\[ (x - 4)(x + 3) = 0 \]
Thus, the solutions are:
\[ x = 4 \quad \text{or} \quad x = -3 \]

#### Case 2: \( x^2 - 12 = -x \)
Rearranging gives:
\[ x^2 + x - 12 = 0 \]
Factoring the quadratic:
\[ (x + 4)(x - 3) = 0 \]
Thus, the solutions are:
\[ x = -4 \quad \text{or} \quad x = 3 \]

### Step 4: Collect All Solutions
From both cases, we have the following solutions:
1. \( x = 4 \)
2. \( x = -3 \)
3. \( x = -4 \)
4. \( x = 3 \)

### Step 5: Check the Sum of Solutions
Now, we can check the sum of the solutions:
\[ 4 + (-3) + (-4) + 3 = 0 \]

### Conclusion
The equation \( 2^{|x^2 - 12|} = \sqrt{e^{|x| \log 4}} \) has **4 real solutions whose sum is 0**.

### Final Answer
The correct option is: **4 real solutions whose sum is 0**.

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