Log3 (x3 -x) - log3 x = log3 3;

To solve the equation \( \log_3 (x^3 - x) - \log_3 x = \log_3 3 \), we can use the properties of logarithms. First, we can combine the logarithms on the left side using the property: \[ \log_a b - \log_a c = \log_a \left( \frac{b}{c} \right) \] Applying this property, we get: \[ \log_3 \left( \frac{x^3 - x}{x} \right) = \log_3 3 \] Simplifying the left side: \[ \frac{x^3 - x}{x} = x^2 - 1 \] So we can rewrite the equation as: \[ \log_3 (x^2 - 1) = \log_3 3 \] Since the logarithms are equal, we set the arguments equal to each other: \[ x^2 - 1 = 3 \] Now, we can solve for \( x \): \[ x^2 = 4 \] Taking the square root of both sides gives us: \[ x = 2 \quad \text{or} \quad x = -2 \] However, we must consider the domain of the original logarithmic expressions. The domain requires that the arguments of all logarithmic functions be positive: 1. \( x > 0 \) for \( \log_3 x \) 2. \( x^3 - x > 0 \) which factors as \( x(x^2 - 1) = x(x - 1)(x + 1) > 0 \). Analyzing the factor \( x(x - 1)(x + 1) > 0 \): - The critical points are \( x = -1, 0, 1 \). - The intervals to test are \( (-\infty, -1) \), \( (-1, 0) \), \( (0, 1) \), and \( (1, \infty) \). Evaluating these intervals shows: - In \( (-\infty, -1) \): all factors are negative, product is negative. - In \( (-1, 0) \): one factor is negative, two are positive, product is negative. - In \( (0, 1) \): \( x > 0 \), \( x - 1 < 0 \), and \( x + 1 > 0 \), product is negative. - In \( (1, \infty) \): all factors are positive, the product is positive. Thus, the only valid solution in the domain is \( x = 2 \). Therefore, the solution to the equation is: \[ \boxed{2} \]