The set of real values of `x` satisfying the equation `|x-1|^(log_3(x^2)-2log_x(9))=(x-1)^7`

Options

A. `1/sqrt3`
B. 1
C. 2
D. 81

Explanation

To solve the equation \( |x-1|^{(\log_3(x^2) - 2\log_x(9))} = (x-1)^7 \), we will follow these steps:

### Step 1: Analyze the domain
The logarithm is defined for \( x > 0 \) and \( x \neq 1 \). Since we are dealing with absolute values and powers, we will also consider the case when \( x > 1 \) to avoid complications with negative bases.

**Hint:** Remember that logarithmic functions are only defined for positive arguments.

### Step 2: Rewrite the equation
We can express the equation as:
\[ |x-1|^{(\log_3(x^2) - 2\log_x(9))} = (x-1)^7 \]
Since \( x > 1 \), we have \( |x-1| = x-1 \). Thus, the equation simplifies to:
\[ (x-1)^{(\log_3(x^2) - 2\log_x(9))} = (x-1)^7 \]

### Step 3: Equate the exponents
Since the bases are the same and \( x-1 > 0 \), we can equate the exponents:
\[ \log_3(x^2) - 2\log_x(9) = 7 \]

### Step 4: Simplify the logarithmic terms
Using the properties of logarithms, we can rewrite \( \log_x(9) \):
\[ \log_x(9) = \frac{\log_3(9)}{\log_3(x)} = \frac{2}{\log_3(x)} \]
Thus, substituting this back into the equation gives:
\[ \log_3(x^2) - 2 \cdot \frac{2}{\log_3(x)} = 7 \]
This simplifies to:
\[ 2\log_3(x) - \frac{4}{\log_3(x)} = 7 \]

### Step 5: Let \( t = \log_3(x) \)
Substituting \( t \) into the equation results in:
\[ 2t - \frac{4}{t} = 7 \]
Multiplying through by \( t \) (assuming \( t \neq 0 \)):
\[ 2t^2 - 4 = 7t \]
Rearranging gives us the quadratic equation:
\[ 2t^2 - 7t - 4 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot (-4)}}{2 \cdot 2} \]
Calculating the discriminant:
\[ t = \frac{7 \pm \sqrt{49 + 32}}{4} = \frac{7 \pm \sqrt{81}}{4} = \frac{7 \pm 9}{4} \]
This gives us two potential solutions:
\[ t_1 = \frac{16}{4} = 4 \quad \text{and} \quad t_2 = \frac{-2}{4} = -\frac{1}{2} \]

### Step 7: Convert back to \( x \)
Recall that \( t = \log_3(x) \):
1. For \( t_1 = 4 \):
\[ x = 3^4 = 81 \]
2. For \( t_2 = -\frac{1}{2} \):
\[ x = 3^{-\frac{1}{2}} = \frac{1}{\sqrt{3}} \]

### Step 8: Determine valid solutions
Since we established that \( x > 1 \), we discard \( x = \frac{1}{\sqrt{3}} \). Thus, the valid solutions are:
\[ x = 2 \quad \text{and} \quad x = 81 \]

### Final Answer
The set of real values of \( x \) satisfying the equation is:
\[ \{2, 81\} \]