For the equation `x^(3/4(logx)^(2)+log_(2)x-5/4)=sqrt2`, which one of the following is true ?

Options

A. Has at least one real solution
B. Has exactly three real solutions
C. Has exactly one irrational solutions
D. Has non-real complex roots

Answer

To solve the equation \( x^{\frac{3}{4}(\log x)^2 + \log_2 x - \frac{5}{4}} = \sqrt{2} \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ x^{\frac{3}{4}(\log x)^2 + \log_2 x - \frac{5}{4}} = \sqrt{2} \]
We know that \( \sqrt{2} = 2^{1/2} \). Thus, we can rewrite the equation as:
\[ x^{\frac{3}{4}(\log x)^2 + \log_2 x - \frac{5}{4}} = 2^{1/2} \]

### Step 2: Take logarithm on both sides
Taking logarithm base 2 on both sides:
\[ \log_2\left(x^{\frac{3}{4}(\log x)^2 + \log_2 x - \frac{5}{4}}\right) = \log_2\left(2^{1/2}\right) \]
Using the property of logarithms, we simplify the left side:
\[ \left(\frac{3}{4}(\log x)^2 + \log_2 x - \frac{5}{4}\right) \log_2 x = \frac{1}{2} \]

### Step 3: Substitute \( \log_2 x \)
Let \( y = \log_2 x \). Then we can rewrite the equation as:
\[ \left(\frac{3}{4}y^2 + y - \frac{5}{4}\right)y = \frac{1}{2} \]
This simplifies to:
\[ \frac{3}{4}y^3 + y^2 - \frac{5}{4}y - \frac{1}{2} = 0 \]

### Step 4: Clear the fraction
To eliminate the fractions, we can multiply the entire equation by 4:
\[ 3y^3 + 4y^2 - 5y - 2 = 0 \]

### Step 5: Factor the polynomial
We can try to factor the cubic polynomial. By using the Rational Root Theorem or synthetic division, we can find the roots. Testing \( y = 1 \):
\[ 3(1)^3 + 4(1)^2 - 5(1) - 2 = 3 + 4 - 5 - 2 = 0 \]
Thus, \( y - 1 \) is a factor. We can perform synthetic division to factor the polynomial:
\[ 3y^3 + 4y^2 - 5y - 2 = (y - 1)(3y^2 + 7y + 2) \]

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

### Step 7: Find \( x \) values
Now we have three values for \( y \):
1. \( y = 1 \) gives \( x = 2^1 = 2 \)
2. \( y = -\frac{1}{3} \) gives \( x = 2^{-\frac{1}{3}} = \frac{1}{\sqrt[3]{2}} \)
3. \( y = -2 \) gives \( x = 2^{-2} = \frac{1}{4} \)

### Conclusion
The solutions for \( x \) are \( 2, \frac{1}{\sqrt[3]{2}}, \frac{1}{4} \). Therefore, the statement that is true is that the equation has at least one real solution, and it also has irrational solutions.