The solution set of inequality
`(1)/(2^(x)-1) gt (1)/(1-2^(x-1))` is

Options

A. `(1,oo)`
B. `(o,log_(2)(4//3))`
C. `(-1,oo)`
D. `(o,log_(2)(4//3)) uu (1,oo)`

Explanation

To solve the inequality

\[ \frac{1}{2^x - 1} > \frac{1}{1 - 2^{x-1}}, \]

we will follow these steps:

### Step 1: Rewrite the Inequality
First, we can rewrite the right-hand side of the inequality:

\[ \frac{1}{1 - 2^{x-1}} = \frac{1}{1 - \frac{2^x}{2}} = \frac{2}{2 - 2^x}. \]

Thus, the inequality becomes:

\[ \frac{1}{2^x - 1} > \frac{2}{2 - 2^x}. \]

### Step 2: Cross-Multiply
Next, we will cross-multiply to eliminate the fractions, keeping in mind that we need to consider the signs of the terms:

\[ (2 - 2^x) > 2(2^x - 1). \]

### Step 3: Simplify the Inequality
Now, simplify the inequality:

\[ 2 - 2^x > 2 \cdot 2^x - 2. \]

This simplifies to:

\[ 2 - 2^x > 2^{x+1} - 2. \]

Rearranging gives:

\[ 2 + 2^x > 2^{x+1}. \]

### Step 4: Factor Out Common Terms
We can factor out \(2^x\):

\[ 2 > 2^{x+1} - 2^x. \]

This simplifies to:

\[ 2 > 2^x(2 - 1) \implies 2 > 2^x. \]

### Step 5: Solve for \(x\)
This inequality can be rewritten as:

\[ 2^x < 2. \]

Taking logarithm base 2 on both sides gives:

\[ x < 1. \]

### Step 6: Consider the Domain
We also need to consider the domain of the original inequality. The expressions \(2^x - 1\) and \(1 - 2^{x-1}\) must not be zero, which means:

1. \(2^x - 1 \neq 0 \implies x \neq 0\)
2. \(1 - 2^{x-1} \neq 0 \implies 2^{x-1} \neq 1 \implies x \neq 1\)

### Final Solution Set
Combining these results, we find that the solution set is:

\[ x < 1 \quad \text{and} \quad x \neq 0. \]

Thus, the solution set is:

\[ (-\infty, 0) \cup (0, 1). \]