The equation ` sqrt(1+log_(x) sqrt27) log_(3) x+1 = 0` has

Options

A. no integral solution
B. one irrational solution
C. two real solutions
D. no prime solution

Explanation

To solve the equation \( \sqrt{1 + \log_{x} \sqrt{27}} \cdot \log_{3} x + 1 = 0 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sqrt{1 + \log_{x} \sqrt{27}} \cdot \log_{3} x + 1 = 0 \]
We can rewrite \( \log_{x} \sqrt{27} \) using the change of base formula:
\[ \log_{x} \sqrt{27} = \frac{\log_{3} \sqrt{27}}{\log_{3} x} \]
Since \( \sqrt{27} = 3^{3/2} \), we have:
\[ \log_{3} \sqrt{27} = \frac{3}{2} \]
Thus, we can rewrite:
\[ \log_{x} \sqrt{27} = \frac{\frac{3}{2}}{\log_{3} x} \]

### Step 2: Substitute into the equation
Substituting this into the original equation gives:
\[ \sqrt{1 + \frac{\frac{3}{2}}{\log_{3} x}} \cdot \log_{3} x + 1 = 0 \]

### Step 3: Set a new variable
Let \( y = \log_{3} x \). Then the equation becomes:
\[ \sqrt{1 + \frac{3}{2y}} \cdot y + 1 = 0 \]

### Step 4: Isolate the square root
Rearranging gives:
\[ \sqrt{1 + \frac{3}{2y}} \cdot y = -1 \]
Squaring both sides:
\[ \left( \sqrt{1 + \frac{3}{2y}} \cdot y \right)^2 = 1 \]
This simplifies to:
\[ (1 + \frac{3}{2y}) y^2 = 1 \]

### Step 5: Expand and rearrange
Expanding gives:
\[ y^2 + \frac{3}{2} y = 1 \]
Multiplying through by 2 to eliminate the fraction:
\[ 2y^2 + 3y - 2 = 0 \]

### Step 6: Factor the quadratic
Now we factor the quadratic:
\[ (2y - 1)(y + 2) = 0 \]
This gives us two cases:
1. \( 2y - 1 = 0 \) leading to \( y = \frac{1}{2} \)
2. \( y + 2 = 0 \) leading to \( y = -2 \)

### Step 7: Solve for \( x \)
Since \( y = \log_{3} x \):
1. From \( y = \frac{1}{2} \):
\[ \log_{3} x = \frac{1}{2} \implies x = 3^{1/2} = \sqrt{3} \]
2. From \( y = -2 \):
\[ \log_{3} x = -2 \implies x = 3^{-2} = \frac{1}{9} \]

### Step 8: Analyze the solutions
The solutions we found are \( x = \sqrt{3} \) and \( x = \frac{1}{9} \).

### Conclusion
- The equation has two real solutions: \( \sqrt{3} \) and \( \frac{1}{9} \).
- The options provided indicate that there are no integral solutions (since both solutions are non-integers) and no prime solutions.