solve the logarithmic equation log_{6}(2x - 6) + log_{6}x = 2 ​

Question

solve the logarithmic equation
log_{6}(2x - 6) + log_{6}x = 2
​

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Nguyệt Ánh 3 years 2021-08-01T22:51:04+00:00 2 Answers 10 views 0

Answers ( )

  1. Jezebel
    0
    2021-08-01T22:52:09+00:00
    Reply

    Answer:

    x=6

    Step-by-step explanation:

    We want to solve the equation:

    \displaystyle \log_6(2x-6)+\log_6x=2

    Recall the property:

    \displaystyle \log_bx+\log_by=\log_b(xy)

    Hence:

    \log_6(x(2x-6))=2

    Next, recall that by the definition of logarithms:

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

    Therefore:

    6^2=x(2x-6)

    Solve for x. Simplify and distribute:

    36=2x^2-6x

    We can divide both sides by two:

    x^2-3x=18

    Subtract 18 from both sides:

    x^2-3x-18=0

    Factor:

    (x-6)(x+3)=0

    Zero Product Property:

    x-6=0\text{ or } x+3=0

    Solve for each case. Hence:

    x=6\text{ or } x=-3

    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:

    \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}

    Hence, x = 6 is indeed a solution.

    Checking x = -3:

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

    Since the second term is undefined, x = -3 is not a solution.

    Therefore, our only solution is x = 6.

  2. thienthi
    0
    2021-08-01T22:52:43+00:00
    Reply

    Answer:

    x = 6

    Step-by-step explanation:

    The given logarithmic equation is ,

    \implies log_{6}(2x - 6) + log_{6}x = 2

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

    \bf \to log_a b + log_a c = log_a {( ac)}

    So we can simplify the LHS and write it as ,

    \implies log_{6} \{ x ( 2x - 6 )\} = 2

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

    \implies log_6 ( 2x^2 - 6x ) = 2

    Again , we know that ,

    \bf \to log_a b = c , a^c = b

    Using this we have ,

    \implies 2x^2 - 6x = 6^2 \\\\\implies 2x^2 - 6x -36 = 0

    Now simplify the quadratic equation ,

    \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

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

    \implies 2x - 6 > 0 \\\\\implies x > 3

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

    \implies \underline{\underline{ x \quad = \quad 6 }}

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