need help solving the equation?

log2 (x)+log2 (x-7)=3

3 Answers

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  • 1 decade ago
    Favorite Answer

    log2(x)+log2(x-7)=3

    x(x-7) = (2)³

    (x - 8) (x + 1) = 0

    x = 8, -1 .... discard -1

    Answer: x = 8

  • 1 decade ago

    log2x + log2(x-7) = 3

    log2 + logx + log2 + log(x-7) = 3

    logx + log(x-7) = 3 - 2log2

    log((x)(x-7)) = 3 - log(2^2)

    10^(3 - log4) = (x)(x-7)

    250 = x^2 - 7x

    x^2 - 7x - 250 = 0

    x = 19.69413474

    x = -12.69413474 (Use Quadratic Formula)

    x cannot be -12.69413474 because it doesn't work in the original equation; you cannot take the log of a negative number. Final answer: x=19.69413474

  • Como
    Lv 7
    1 decade ago

    Let log be log to base 2

    log [ ( x ) ( x - 7 ) ] = 3

    [ ( x ) ( x - 7 ) ] = 2 ³

    x² - 7x - 8 = 0

    ( x - 8 ) ( x + 1 ) = 0

    x = 8 is acceptable

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