Log3(x-2)+log3(x+6)=2

To solve the equation ${\log }_3(x-2)+{\log }_3(x+6)=2$, follow these logical steps: 1. Determine the Domain (ODZ) The argument of a logarithm must be strictly greater than zero. Therefore, we must satisfy the following system of inequalities:

• $x-2>0⟹x>2$ $x+6>0⟹x>-6$

The intersection of these conditions is $x>2$. Any potential solution must fall within this range. 2. Apply Logarithmic Properties Using the product rule for logarithms, ${\log }_b\left(m\right)+{\log }_b\left(n\right)={\log }_b(m\cdot n)$, we can combine the left side: ${\log }_3\left(\right(x-2\left)\right(x+6\left)\right)=2$3. Convert to Exponential Form By the definition of a logarithm ( ${\log }_b\left(a\right)=c⟺b^c=a$), we rewrite the equation: $(x-2)(x+6)=3^2$ $(x-2)(x+6)=9$4. Solve the Quadratic Equation Expand the brackets and move all terms to one side: $x^2+6x-2x-12=9$ $x^2+4x-12-9=0$ $x^2+4x-21=0$Using the quadratic formula or Vieta's formulas:

• Sum of roots: $x_1+x_2=-4$ Product of roots: $x_1\cdot x_2=-21$

The roots are:

• $x_1=3$
• $x_2=-7$

5. Verify Against the Domain We check each root against our initial condition ( $x>2$):

• $x=3$: $3>2$ (Valid) $x=-7$: $-7<2$ (Invalid/Extraneous)

Answer: $x=3$ Would you like me to solve another logarithmic equation or demonstrate how to check the result by substituting it back into the original equation?