Solve the following equations.
(x) `log_ax=x`, where `a=x^(log_ax)`

Answer

To solve the equation \( \log_a x = x \) where \( a = x^{\log_a x} \), we can follow these steps:

### Step 1: Write down the equations
We have two equations:
1. \( \log_a x = x \) (Equation 1)
2. \( a = x^{\log_a x} \) (Equation 2)

### Step 2: Substitute Equation 2 into Equation 1
From Equation 2, substitute \( a \) into Equation 1:
\[ \log_{(x^{\log_a x})} x = x \]

### Step 3: Use the property of logarithms
Using the property of logarithms, \( \log_{b^m} n = \frac{1}{m} \log_b n \), we can rewrite the left-hand side:
\[ \frac{1}{\log_a x} \log x = x \]

### Step 4: Simplify the equation
Since \( \log_a x = x \) from Equation 1, we can replace \( \log_a x \) in the equation:
\[ \frac{1}{x} \log x = x \]

### Step 5: Rearranging the equation
Multiply both sides by \( x \):
\[ \log x = x^2 \]

### Step 6: Set up the final equation
We now have the equation:
\[ x^2 - \log x = 0 \]

### Step 7: Solve for \( x \)
This can be rewritten as:
\[ x^2 = \log x \]
To find the values of \( x \), we can analyze the function \( f(x) = x^2 - \log x \).

### Step 8: Find critical points
We can find where \( f(x) = 0 \) by testing values:
1. For \( x = 1 \):
\[ f(1) = 1^2 - \log 1 = 1 - 0 = 1 \quad (\text{not a solution}) \]
2. For \( x = 0.5 \):
\[ f(0.5) = (0.5)^2 - \log(0.5) = 0.25 + 0.693 \approx 0.943 \quad (\text{not a solution}) \]
3. For \( x = 2 \):
\[ f(2) = 2^2 - \log(2) = 4 - 0.301 \approx 3.699 \quad (\text{not a solution}) \]

### Step 9: Analyze the function
Since \( \log x \) grows slower than \( x^2 \) for \( x > 1 \) and \( \log x \) is negative for \( 0 < x < 1 \), we can conclude that the only solution is when \( x = 1 \).

### Final Solution
Thus, the only valid solution is:
\[ \boxed{1} \]