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: Take logarithm on both sides
We start by taking the logarithm (base 3) of both sides of the equation:
\[ \log_3\left(x^{\left(\frac{3}{4}(\log_3 x)^2 + \log_3 x - \frac{5}{4}\right)}\right) = \log_3(\sqrt{3}) \]

### Step 2: Simplify the left side using the power rule of logarithms
Using the property of logarithms that states \( \log_b(a^c) = c \cdot \log_b(a) \), we can simplify the left side:
\[ \left(\frac{3}{4}(\log_3 x)^2 + \log_3 x - \frac{5}{4}\right) \cdot \log_3 x = \log_3(3^{1/2}) \]
Since \( \log_3(\sqrt{3}) = \frac{1}{2} \).

### Step 3: Set up the equation
Now we can write the equation as:
\[ \left(\frac{3}{4}(\log_3 x)^2 + \log_3 x - \frac{5}{4}\right) \cdot \log_3 x = \frac{1}{2} \]

### Step 4: Let \( y = \log_3 x \)
Let \( y = \log_3 x \). Then the equation becomes:
\[ \frac{3}{4}y^2 + y - \frac{5}{4} = \frac{1}{2y} \]

### Step 5: Multiply through by \( 4y \) to eliminate the fraction
Multiplying both sides by \( 4y \) gives:
\[ 3y^3 + 4y^2 - 5y = 2 \]

### Step 6: Rearranging the equation
Rearranging this gives us a cubic equation:
\[ 3y^3 + 4y^2 - 5y - 2 = 0 \]

### Step 7: Use the Rational Root Theorem to find possible rational roots
Testing \( y = 1 \):
\[ 3(1)^3 + 4(1)^2 - 5(1) - 2 = 3 + 4 - 5 - 2 = 0 \]
Thus, \( y = 1 \) is a root.

### Step 8: Perform polynomial long division
Now we will divide \( 3y^3 + 4y^2 - 5y - 2 \) by \( y - 1 \):
1. Divide \( 3y^3 \) by \( y \) to get \( 3y^2 \).
2. Multiply \( 3y^2 \) by \( y - 1 \) to get \( 3y^3 - 3y^2 \).
3. Subtract to get \( 7y^2 - 5y - 2 \).
4. Divide \( 7y^2 \) by \( y \) to get \( 7y \).
5. Multiply \( 7y \) by \( y - 1 \) to get \( 7y^2 - 7y \).
6. Subtract to get \( 2y - 2 \).
7. Divide \( 2y \) by \( y \) to get \( 2 \).
8. Multiply \( 2 \) by \( y - 1 \) to get \( 2y - 2 \).
9. Subtract to get \( 0 \).

Thus, we have:
\[ 3y^3 + 4y^2 - 5y - 2 = (y - 1)(3y^2 + 7y + 2) \]

### Step 9: Solve the quadratic equation
Now we need to solve \( 3y^2 + 7y + 2 = 0 \) using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 3 \cdot 2}}{2 \cdot 3} \]
\[ y = \frac{-7 \pm \sqrt{49 - 24}}{6} = \frac{-7 \pm \sqrt{25}}{6} = \frac{-7 \pm 5}{6} \]
This gives us two solutions:
1. \( y = \frac{-2}{6} = -\frac{1}{3} \)
2. \( y = \frac{-12}{6} = -2 \)

### Step 10: Convert back to \( x \)
Now we convert back to \( x \):
1. From \( y = 1 \): \( \log_3 x = 1 \) implies \( x = 3^1 = 3 \)
2. From \( y = -\frac{1}{3} \): \( \log_3 x = -\frac{1}{3} \) implies \( x = 3^{-\frac{1}{3}} = \frac{1}{\sqrt[3]{3}} \)
3. From \( y = -2 \): \( \log_3 x = -2 \) implies \( x = 3^{-2} = \frac{1}{9} \)

### Final Values of \( x \)
Thus, the values of \( x \) are:
- \( x = 3 \) (rational)
- \( x = \frac{1}{\sqrt[3]{3}} \) (irrational)
- \( x = \frac{1}{9} \) (rational)

### Conclusion
The nature of the values of \( x \) is:
- One rational value: \( 3 \)
- One irrational value: \( \frac{1}{\sqrt[3]{3}} \)
- One rational value: \( \frac{1}{9} \)