If `x_1` and `x_2` `(x_2gtx_1)` are the integral solutions of the equation
`(log_5x)^2+log_(5x)(5/x)=1`, , the value of `|x_2-4x_1|` is

Answer

To solve the equation \((\log_5 x)^2 + \log_{5x} \left(\frac{5}{x}\right) = 1\), we will follow these steps:

### Step 1: Change the base of the logarithm
We can rewrite \(\log_{5x} \left(\frac{5}{x}\right)\) using the change of base formula:
\[ \log_{5x} \left(\frac{5}{x}\right) = \frac{\log_5 \left(\frac{5}{x}\right)}{\log_5 (5x)} \]
This gives us:
\[ \log_{5x} \left(\frac{5}{x}\right) = \frac{\log_5 5 - \log_5 x}{\log_5 5 + \log_5 x} = \frac{1 - \log_5 x}{1 + \log_5 x} \]

### Step 2: Substitute \(y = \log_5 x\)
Let \(y = \log_5 x\). The equation now becomes:
\[ y^2 + \frac{1 - y}{1 + y} = 1 \]

### Step 3: Simplify the equation
Now, we can simplify:
\[ y^2 + \frac{1 - y}{1 + y} - 1 = 0 \]
This can be rewritten as:
\[ y^2 + \frac{1 - y - (1 + y)}{1 + y} = 0 \]
\[ y^2 + \frac{-2y}{1 + y} = 0 \]
Multiplying through by \(1 + y\) (assuming \(y \neq -1\)):
\[ y^2(1 + y) - 2y = 0 \]
\[ y^3 - 2y + y^2 = 0 \]
\[ y^3 + y^2 - 2y = 0 \]

### Step 4: Factor the equation
Factoring out \(y\):
\[ y(y^2 + y - 2) = 0 \]
Now factor the quadratic:
\[ y^2 + y - 2 = (y - 1)(y + 2) \]
Thus, we have:
\[ y(y - 1)(y + 2) = 0 \]

### Step 5: Solve for \(y\)
The solutions for \(y\) are:
\[ y = 0, \quad y = 1, \quad y = -2 \]

### Step 6: Find \(x\)
Converting back to \(x\):
1. If \(y = 0\), then \(x = 5^0 = 1\).
2. If \(y = 1\), then \(x = 5^1 = 5\).
3. If \(y = -2\), then \(x = 5^{-2} = \frac{1}{25}\) (not an integer).

Thus, the integral solutions are \(x_1 = 1\) and \(x_2 = 5\) (where \(x_2 > x_1\)).

### Step 7: Calculate \(|x_2 - 4x_1|\)
Now we need to find:
\[ |x_2 - 4x_1| = |5 - 4 \cdot 1| = |5 - 4| = |1| = 1 \]

### Final Answer
The value of \(|x_2 - 4x_1|\) is \(1\).

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