Find the domain `f(x)=log_(100x)((2 log_(10) x+1)/-x)`

Answer

To find the domain of the function \( f(x) = \log_{100x}\left(\frac{2 \log_{10} x + 1}{-x}\right) \), we need to ensure that both the base and the argument of the logarithm are valid.

### Step 1: Conditions for the base of the logarithm
The base of the logarithm \( 100x \) must satisfy two conditions:
1. \( 100x > 0 \)
2. \( 100x \neq 1 \)

From the first condition:
\[ 100x > 0 \implies x > 0 \]

From the second condition:
\[ 100x \neq 1 \implies x \neq \frac{1}{100} \]

### Step 2: Conditions for the argument of the logarithm
The argument of the logarithm \( \frac{2 \log_{10} x + 1}{-x} \) must be greater than zero:
\[ \frac{2 \log_{10} x + 1}{-x} > 0 \]
This inequality can be split into two cases based on the sign of the numerator and denominator.

#### Case 1: \( -x > 0 \) (which implies \( x < 0 \))
This case is not valid since we already established that \( x > 0 \).

#### Case 2: \( -x < 0 \) (which implies \( x > 0 \))
In this case, we need:
\[ 2 \log_{10} x + 1 > 0 \implies 2 \log_{10} x > -1 \implies \log_{10} x > -\frac{1}{2} \]
Taking the antilogarithm:
\[ x > 10^{-\frac{1}{2}} = \frac{1}{\sqrt{10}} \]

### Step 3: Combine the conditions
Now we combine the conditions we found:
1. \( x > 0 \)
2. \( x \neq \frac{1}{100} \)
3. \( x > \frac{1}{\sqrt{10}} \)

Since \( \frac{1}{\sqrt{10}} \approx 0.316 \) and \( \frac{1}{100} = 0.01 \), we can summarize the domain:
- The values of \( x \) must be greater than \( \frac{1}{\sqrt{10}} \) and cannot include \( \frac{1}{100} \).

### Step 4: Final domain
Thus, the domain of \( f(x) \) is:
\[ \boxed{( \frac{1}{\sqrt{10}}, \infty )} \]