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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]