solve the logarithmic equation [tex] log_{6}(2x – 6) + log_{6}x = 2[/tex] ​

Question

solve the logarithmic equation
[tex] log_{6}(2x – 6) + log_{6}x = 2[/tex]

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Nguyệt Ánh 1 year 2021-08-01T22:51:04+00:00 2 Answers 5 views 0

Answers ( )

    0
    2021-08-01T22:52:09+00:00

    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.

    0
    2021-08-01T22:52:43+00:00

    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]

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