Lg(x^2-9)-lg(x+3)=0

To solve the equation \( \log(x^2 - 9) - \log(x + 3) = 0 \), we can use the properties of logarithms.

1. Combine the logarithms: \[ \log\left(\frac{x^2 - 9}{x + 3}\right) = 0 \]

2. Exponentiate both sides: Since \( \log(a) = 0 \) implies \( a = 1 \): \[ \frac{x^2 - 9}{x + 3} = 1 \]

3. Multiply both sides by \( x + 3 \) (assuming \( x + 3 \neq 0 \)): \[ x^2 - 9 = x + 3 \]

4. Rearrange the equation: \[ x^2 - x - 12 = 0 \]

5. Factor the quadratic equation: \[ (x - 4)(x + 3) = 0 \]

6. Set each factor to zero: \[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \] \[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \]

7. Check for extraneous solutions: We need to ensure that the arguments of the logarithms are positive:

   • For \( x = 4\):
     • \( x^2 - 9 = 4^2 - 9 = 16 - 9 = 7 > 0 \)
     • \( x + 3 = 4 + 3 = 7 > 0 \)
   • For \( x = -3\):
     • \( x^2 - 9 = (-3)^2 - 9 = 9 - 9 = 0 \), which is not valid for logarithms (log of 0 is undefined).
     • \( x + 3 = -3 + 3 = 0 \), which is also not valid.

Therefore, the only valid solution is: \[ \boxed{4} \]