If `log_(5)(6+(2)/(x))+log_((1//5))(1+(x)/(10))le1`, then `x` lies in

Options

A. `(1-sqrt(5),1)`
B. `(1,1+sqrt(5))`
C. `(1-sqrt(5),1+sqrt(5))`
D. `(-oo,1-sqrt(5)]uu[1+sqrt(5),oo)`

Answer

To solve the inequality \( \log_{5}\left(6 + \frac{2}{x}\right) + \log_{\frac{1}{5}}\left(1 + \frac{x}{10}\right) \leq 1 \), we will follow these steps:

### Step 1: Rewrite the logarithm
Using the property of logarithms, we know that \( \log_{\frac{1}{5}}(a) = -\log_{5}(a) \). Therefore, we can rewrite the inequality as:

\[ \log_{5}\left(6 + \frac{2}{x}\right) - \log_{5}\left(1 + \frac{x}{10}\right) \leq 1 \]

### Step 2: Combine the logarithms
Using the property \( \log_{a}(b) - \log_{a}(c) = \log_{a}\left(\frac{b}{c}\right) \), we can combine the logarithms:

\[ \log_{5}\left(\frac{6 + \frac{2}{x}}{1 + \frac{x}{10}}\right) \leq 1 \]

### Step 3: Exponentiate both sides
To eliminate the logarithm, we exponentiate both sides. Since the base \( 5 \) is greater than \( 1 \), the inequality direction remains the same:

\[ \frac{6 + \frac{2}{x}}{1 + \frac{x}{10}} \leq 5 \]

### Step 4: Cross-multiply
Cross-multiplying gives us:

\[ 6 + \frac{2}{x} \leq 5\left(1 + \frac{x}{10}\right) \]

Expanding the right side:

\[ 6 + \frac{2}{x} \leq 5 + \frac{5x}{10} \]

### Step 5: Simplify the inequality
This simplifies to:

\[ 6 + \frac{2}{x} \leq 5 + \frac{x}{2} \]

Subtract \( 5 \) from both sides:

\[ 1 + \frac{2}{x} \leq \frac{x}{2} \]

### Step 6: Eliminate the fraction
To eliminate the fraction, multiply through by \( 2x \) (assuming \( x > 0 \)):

\[ 2x + 4 \leq x^2 \]

### Step 7: Rearrange the inequality
Rearranging gives:

\[ x^2 - 2x - 4 \geq 0 \]

### Step 8: Solve the quadratic inequality
To solve \( x^2 - 2x - 4 = 0 \), we use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \]

Calculating the discriminant:

\[ x = \frac{2 \pm \sqrt{4 + 16}}{2} = \frac{2 \pm \sqrt{20}}{2} = \frac{2 \pm 2\sqrt{5}}{2} = 1 \pm \sqrt{5} \]

### Step 9: Determine the intervals
The roots are \( 1 - \sqrt{5} \) and \( 1 + \sqrt{5} \). The quadratic opens upwards, so the solution to the inequality \( x^2 - 2x - 4 \geq 0 \) is:

\[ x \leq 1 - \sqrt{5} \quad \text{or} \quad x \geq 1 + \sqrt{5} \]

### Final Solution
Thus, the solution set for \( x \) is:

\[ (-\infty, 1 - \sqrt{5}] \cup [1 + \sqrt{5}, \infty) \]