Algebra 1

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solve: log8(w-6)=2-log8(w+15)
note: the 8 is to be little at the bottom right corner of log

  • Algebra 1 -

    since 64 = 8^2, then using base-8 logs, we have

    log(w-6) = log(64)-log(w+15)
    (w-6) = 64/(w+15)
    (w-6)(w+15) = 64
    w^2 + 9w - 154 = 0
    w = 1/2 (-9±√697)
    scrap the negative value (why?)

  • Algebra 1 -

    because the answer has got to be a real number which is not a negative number

  • Algebra 1 -

    how do you get 1/2 as a solution?

  • Algebra 1 -

    So technically there would be no solution right?

  • Algebra 1 -

    So, technically, the solution is

    x = 1/2 (-9+√697) = 8.70

    where does the 1/2 come from? Think back ... back ... back to Algebra I and the quadratic formula.

  • Algebra 1 -

    the quadratic formula is -b+-radical b^2-4ac/a

  • Algebra 1 -

    SO HOW DO YOU GET 1/2 FORM THAT?

  • Algebra 1 -

    Britt, when Steve and I went to school, the quadratic formula used to be
    (-b ± √(b^2 -4ac)/(2a) or
    (1/2) (-b ± √(b^2 -4ac)/a , thus the 1/2

    I was not aware of any recent changes to the formula.

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