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1 July, 04:10

Solve the equation:

2logx-log (x-2) = 2log3

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Answers (1)
  1. 1 July, 05:40
    0
    x=6 x=3

    Step-by-step explanation:

    2logx-log (x-2) = 2log3

    We know that a log b = log b^a

    logx^2-log (x-2) = log3^2

    logx^2-log (x-2) = log9

    subtract log 9 from each side

    logx^2-log (x-2) - log9=0

    We know that log a - log b = log (a/b)

    log (x^2/9 (x-2)) = 0

    Raise each side to the power of 10 to get rid of the log

    10 ^log (x^2/9 (x-2)) = 10 ^0

    10^ log cancels and 10^0 = 1

    (x^2/9 (x-2)) = 1

    Multiply each side by 9 (x-2)

    x^2 = 9 (x-2)

    Subtract 9 (x-2) from each side

    x^2 - 9 (x-2) = 9 (x-2) - 9 (x-2)

    x^2 - 9 (x-2) = 0

    Distribute

    x^2 - 9x + 18 = 0

    Factor

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

    Using the zero product property

    x-6 = 0 x-3 = 0

    x=6 x=3
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