Given:
The equation is,
[tex]2\log _3x-\log _3(x-2)=2[/tex]Explanation:
Simplify the equation by using logarthimic property.
[tex]\begin{gathered} 2\log _3x-\log _3(x-2)=2 \\ \log _3x^2-\log _3(x-2)=2_{}\text{ \lbrack{}log(a)-log(b) = log(a/b)\rbrack} \\ \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \end{gathered}[/tex]Simplify further.
[tex]\begin{gathered} \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \\ \frac{x^2}{x-2}=3^2 \\ x^2=9(x-2) \\ x^2-9x+18=0 \end{gathered}[/tex]Solve the quadratic equation for x.
[tex]\begin{gathered} x^2-6x-3x+18=0 \\ x(x-6)-3(x-6)=0 \\ (x-6)(x-3)=0 \end{gathered}[/tex]From the above equation (x - 6) = 0 or (x - 3) = 0.
For (x - 6) = 0,
[tex]\begin{gathered} x-6=0 \\ x=6 \end{gathered}[/tex]For (x - 3) = 0,
[tex]\begin{gathered} x-3=0 \\ x=3 \end{gathered}[/tex]The values of x from solving the equations are x = 3 and x = 6.
Substitute the values of x in the equation to check answers are valid or not.
For x = 3,
[tex]\begin{gathered} 2\log _3(3^{})-\log _3(3-2)=2 \\ 2\log _33-\log _31=2 \\ 2\cdot1-0=2 \\ 2=2 \end{gathered}[/tex]Equation satisfy for x = 3. So x = 3 is valid value of x.
For x = 6,
[tex]\begin{gathered} 2\log _36-\log _3(6-2)=2 \\ 2\log _36-\log _34=2 \\ \log _3(6^2)-\log _34=2 \\ \log _3(\frac{36}{4})=2 \\ \log _39=2 \\ \log _3(3^2)=2 \\ 2\log _33=2 \\ 2=2 \end{gathered}[/tex]Equation satifies for x = 6.
Thus values of x for equation are x = 3 and x = 6.
