Equations `log_(2)x(x-1)=1` and `log_(2)x+log_(2)(x-1)=1`

Answer

To solve the equations \( \log_2(x(x-1)) = 1 \) and \( \log_2(x) + \log_2(x-1) = 1 \), we will break down the steps for each equation.

### Step 1: Solve the first equation \( \log_2(x(x-1)) = 1 \)

1. **Convert the logarithmic equation to its exponential form**:
\[ x(x-1) = 2^1 \]
This simplifies to:
\[ x(x-1) = 2 \]

2. **Rearrange the equation**:
\[ x^2 - x - 2 = 0 \]

3. **Factor the quadratic equation**:
\[ (x-2)(x+1) = 0 \]

4. **Find the solutions**:
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]

5. **Check the validity of the solutions**:
- For \( x = 2 \):
\[ x(x-1) = 2(2-1) = 2 \quad \text{(valid)} \]
- For \( x = -1 \):
\[ x(x-1) = -1(-1-1) = 2 \quad \text{(not valid since x must be positive)} \]

Thus, the only valid solution for the first equation is \( x = 2 \).

### Step 2: Solve the second equation \( \log_2(x) + \log_2(x-1) = 1 \)

1. **Use the property of logarithms**:
\[ \log_2(x(x-1)) = 1 \]

2. **Convert the logarithmic equation to its exponential form**:
\[ x(x-1) = 2^1 \]
This simplifies to:
\[ x(x-1) = 2 \]

3. **Rearrange the equation**:
\[ x^2 - x - 2 = 0 \]

4. **Factor the quadratic equation**:
\[ (x-2)(x+1) = 0 \]

5. **Find the solutions**:
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]

6. **Check the validity of the solutions**:
- For \( x = 2 \):
\[ \log_2(2) + \log_2(1) = 1 + 0 = 1 \quad \text{(valid)} \]
- For \( x = -1 \):
\[ \log_2(-1) \quad \text{(not valid since x must be positive)} \]

Thus, the only valid solution for the second equation is also \( x = 2 \).

### Conclusion

- For the first equation, the valid solution is \( x = 2 \).
- For the second equation, the valid solution is also \( x = 2 \).
- The solution \( x = -1 \) is not valid in both cases.