Next, recall that by the definition of logarithms:

[tex]\displaystyle \log_b(a)=c\text{ if and only if } b^c=a[/tex]

Therefore:

[tex]6^2=x(2x-6)[/tex]

Solve for x. Simplify and distribute:

[tex]36=2x^2-6x[/tex]

We can divide both sides by two:

[tex]x^2-3x=18[/tex]

Subtract 18 from both sides:

[tex]x^2-3x-18=0[/tex]

Factor:

[tex](x-6)(x+3)=0[/tex]

Zero Product Property:

[tex]x-6=0\text{ or } x+3=0[/tex]

Solve for each case. Hence:

[tex]x=6\text{ or } x=-3[/tex]

Next, we must check the solutions for extraneous solutions. To do so, we can simply substitute the solutions back into the original equations and examine its validity.

Answer:[tex]x=6[/tex]

Step-by-step explanation:We want to solve the equation:

[tex]\displaystyle \log_6(2x-6)+\log_6x=2[/tex]

Recall the property:

[tex]\displaystyle \log_bx+\log_by=\log_b(xy)[/tex]

Hence:

[tex]\log_6(x(2x-6))=2[/tex]

Next, recall that by the definition of logarithms:

[tex]\displaystyle \log_b(a)=c\text{ if and only if } b^c=a[/tex]

Therefore:

[tex]6^2=x(2x-6)[/tex]

Solve for

x. Simplify and distribute:[tex]36=2x^2-6x[/tex]

We can divide both sides by two:

[tex]x^2-3x=18[/tex]

Subtract 18 from both sides:

[tex]x^2-3x-18=0[/tex]

Factor:

[tex](x-6)(x+3)=0[/tex]

Zero Product Property:

[tex]x-6=0\text{ or } x+3=0[/tex]

Solve for each case. Hence:

[tex]x=6\text{ or } x=-3[/tex]

Next, we must check the solutions for extraneous solutions. To do so, we can simply substitute the solutions back into the original equations and examine its validity.

Checking

x= 6:[tex]\displaystyle \begin{aligned} \log_{6}(2(6)-6)+\log_{6}6&\stackrel{?}{=} 2 \\ \\ \log_6(12-6)+(1)&\stackrel{?}{=}2 \\ \\ \log_6(6)+1&\stackrel{?}{=}2 \\ \\ 1+1=2&\stackrel{\checkmark}{=}2\end{aligned}[/tex]

Hence,

x= 6 is indeed a solution.Checking

x= -3:[tex]\displaystyle\begin{aligned} \log_6(2(-3)-6) + \underbrace{\log_6-3}_{\text{und.}} &\stackrel{?}{=} 2\\ \\ \end{aligned}[/tex]

Since the second term is undefined,

x= -3 is not a solution.Therefore, our only solution is

x= 6.Answer:x = 6

Step-by-step explanation:The given logarithmic equation is ,

[tex]\implies log_{6}(2x – 6) + log_{6}x = 2[/tex]

We can notice that the bases of both logarithm is same . So we can use a property of log as ,

[tex]\bf \to log_a b + log_a c = log_a {( ac)} [/tex]

So we can simplify the LHS and write it as ,

[tex]\implies log_{6} \{ x ( 2x – 6 )\} = 2 [/tex]

Now simplify out x(2x – 6 ) . We get ,

[tex]\implies log_6 ( 2x^2 – 6x ) = 2 [/tex]

Again , we know that ,

[tex]\bf \to log_a b = c , a^c = b [/tex]

Using this we have ,

[tex]\implies 2x^2 – 6x = 6^2 \\\\\implies 2x^2 – 6x -36 = 0 [/tex]

Now simplify the quadratic equation ,

[tex]\implies x^2 – 3x – 18 = 0 \\\\\implies x^2 -6x + 3x -18=0\\\\\implies x( x -6) +3( x – 6 ) = 0 \\\\\implies (x-6)(x+3) = 0 \\\\\implies x = 6 , -3 [/tex]

Since logarithms are not defined for negative numbers or zero , therefore ,

[tex]\implies 2x – 6 > 0 \\\\\implies x > 3 [/tex]

Therefore the equation is not defined at x = -3 . Hence the possible value of x is 6 .

[tex]\implies \underline{\underline{ x \quad = \quad 6 }}[/tex]