Log3(x^2-5x+4)-log3(x-4)=2

To solve the equation \( \log_3(x^2 - 5x + 4) - \log_3(x - 4) = 2 \), we can use the properties of logarithms to combine the two logarithmic terms on the left-hand side. Using the property \( \log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right) \), we can rewrite the equation as: \[ \log_3\left(\frac{x^2 - 5x + 4}{x - 4}\right) = 2 \] Next, we can convert this logarithmic equation to its exponential form. Recall that if \( \log_b(a) = c \), then \( a = b^c \). Thus, we have: \[ \frac{x^2 - 5x + 4}{x - 4} = 3^2 \] \[ \frac{x^2 - 5x + 4}{x - 4} = 9 \] Now, we can multiply both sides by \( x - 4 \) (assuming \( x - 4 \neq 0 \)), giving us: \[ x^2 - 5x + 4 = 9(x - 4) \] Expanding the right-hand side: \[ x^2 - 5x + 4 = 9x - 36 \] Now, we can move all terms to one side to form a quadratic equation: \[ x^2 - 5x + 4 - 9x + 36 = 0 \] \[ x^2 - 14x + 40 = 0 \] Next, we can factor this quadratic: \[ (x - 10)(x - 4) = 0 \] Setting each factor equal to zero gives us the potential solutions: \[ x - 10 = 0 \quad \Rightarrow \quad x = 10 \] \[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \] We must check for extraneous solutions since the logarithm is defined only for positive arguments. 1. Check \( x = 10 \): - \( x^2 - 5x + 4 = 10^2 - 5(10) + 4 = 100 - 50 + 4 = 54 \) (positive) - \( x - 4 = 10 - 4 = 6 \) (positive) 2. Check \( x = 4 \): - \( x - 4 = 4 - 4 = 0 \) (not allowed since log can't take zero) Thus, the only valid solution is: \[ \boxed{10} \]