379 1) \( \lg \left(x^{2}-2 x\right)=\lg 30-1 \)

To solve the equation \( \lg \left(x^{2}-2 x\right) = \lg 30 - 1 \), we can first rewrite the right side. Recall that \( \lg a - b = \lg(a/10^b) \), thus: \[ \lg \left(x^{2}-2 x\right) = \lg \left(\frac{30}{10^1}\right) = \lg 3 \] Since the logs are equal, we can drop the logarithm (assuming the arguments are positive) to set the insides equal to each other: \[ x^{2} - 2x = 3 \] Rearranging gives us the quadratic equation: \[ x^{2} - 2x - 3 = 0 \] Now we can factor this equation: \[ (x - 3)(x + 1) = 0 \] Setting each factor to zero gives us two potential solutions: \[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \] \[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \] However, we need to check these solutions in the original logarithmic equation to ensure they don't yield negative arguments: For \( x = 3 \): \[ x^{2} - 2x = 3^{2} - 2 \cdot 3 = 9 - 6 = 3 \quad \text{(valid)} \] For \( x = -1 \): \[ x^{2} - 2x = (-1)^{2} - 2 \cdot (-1) = 1 + 2 = 3 \quad \text{(valid)} \] Thus, both solutions are valid, and the final solutions are: \[ x = 3 \quad \text{and} \quad x = -1 \]