If `x^({(3)/(4)("log"_(3)x)^(2) + ("log"_(3)x)-(5)/(4)}) = sqrt(3)`, then x has

Options

A. all integral values
B. two integral values and one irrational values
C. all irrational values
D. two rational values and an irrational value

Explanation

To solve the equation \( x^{\left(\frac{3}{4}(\log_3 x)^2 + \log_3 x - \frac{5}{4}\right)} = \sqrt{3} \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ x^{\left(\frac{3}{4}(\log_3 x)^2 + \log_3 x - \frac{5}{4}\right)} = \sqrt{3} \]
We can express \(\sqrt{3}\) as \(3^{1/2}\):
\[ x^{\left(\frac{3}{4}(\log_3 x)^2 + \log_3 x - \frac{5}{4}\right)} = 3^{1/2} \]

### Step 2: Take logarithm on both sides
Taking logarithm base 3 on both sides gives:
\[ \log_3\left(x^{\left(\frac{3}{4}(\log_3 x)^2 + \log_3 x - \frac{5}{4}\right)}\right) = \log_3(3^{1/2}) \]
Using the property of logarithms, we can simplify the left side:
\[ \left(\frac{3}{4}(\log_3 x)^2 + \log_3 x - \frac{5}{4}\right) \cdot \log_3 x = \frac{1}{2} \]

### Step 3: Let \(y = \log_3 x\)
Let \(y = \log_3 x\). Then, we can rewrite the equation as:
\[ \left(\frac{3}{4}y^2 + y - \frac{5}{4}\right) y = \frac{1}{2} \]
Expanding this gives:
\[ \frac{3}{4}y^3 + y^2 - \frac{5}{4}y = \frac{1}{2} \]

### Step 4: Multiply through by 4 to eliminate fractions
Multiplying the entire equation by 4 to eliminate the fractions:
\[ 3y^3 + 4y^2 - 5y = 2 \]
Rearranging gives:
\[ 3y^3 + 4y^2 - 5y - 2 = 0 \]

### Step 5: Solve the cubic equation
Now we need to solve the cubic equation \(3y^3 + 4y^2 - 5y - 2 = 0\). We can try to find rational roots using the Rational Root Theorem or synthetic division.

Testing \(y = 1\):
\[ 3(1)^3 + 4(1)^2 - 5(1) - 2 = 3 + 4 - 5 - 2 = 0 \]
Thus, \(y = 1\) is a root.

### Step 6: Factor the cubic polynomial
We can factor \(3y^3 + 4y^2 - 5y - 2\) as \((y - 1)(3y^2 + 7y + 2)\).

### Step 7: Solve the quadratic equation
Now we solve the quadratic \(3y^2 + 7y + 2 = 0\) using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 3 \cdot 2}}{2 \cdot 3} = \frac{-7 \pm \sqrt{49 - 24}}{6} = \frac{-7 \pm \sqrt{25}}{6} = \frac{-7 \pm 5}{6} \]
This gives us:
\[ y = \frac{-2}{6} = -\frac{1}{3} \quad \text{or} \quad y = \frac{-12}{6} = -2 \]

### Step 8: Find \(x\)
Recall \(y = \log_3 x\):
1. For \(y = 1\), \(x = 3^1 = 3\).
2. For \(y = -\frac{1}{3}\), \(x = 3^{-\frac{1}{3}} = \frac{1}{\sqrt[3]{3}}\).
3. For \(y = -2\), \(x = 3^{-2} = \frac{1}{9}\).

### Final Answer
Thus, the values of \(x\) are:
\[ x = 3, \quad x = \frac{1}{\sqrt[3]{3}}, \quad x = \frac{1}{9} \]