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Math Help - Logs eek!

  1. #1
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    Logs eek!

    log(x-16)+log(x-64)=2


    100=X^2-8+1024

    X^2-8+924
    (x- ) (x- )


    this is how far i got. helP!!
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  2. #2
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    Hello, karleewil!

    I don't understand your steps . . .


    \log(x-16)+\log(x-64)\:=\:2
    Note that: . x \,>\,64


    We have: . \log(x-16)(x-64) \:=\:10\quad\Rightarrow\quad \log(x^2-80x + 1024)\:=\:2

    . . x^2-80x + 1024 \:=\:100 \quad\Rightarrow\quad x^2-80x+924\:=\:0


    Factor: . (x - 14)(x - 66) \:=\:0 \quad\Rightarrow\quad x\:=\:14,\:66


    The only acceptable root is: . x \:=\:66

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  3. #3
    Senior Member nikhil's Avatar
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    Lightbulb Check it out

    log(x-16)+log(x-64)=2
    2=log 100 (base is 10)
    therfor log(x-16)+log(x-64)=log(100)
    recall that loga+logb=logab
    therfor log((x-16)(x-64))=log100
    or (x-16)(x-64)=100
    by solving them you will get x=14 and x=66
    we will neglect x=14 as it do not satisfy the domain of given equation(it makes log negative i.e log(-2) and log (-50))
    so the final answer is x=66
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  4. #4
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    Quote Originally Posted by karleewil View Post
    log(x-16)+log(x-64)=2


    100=X^2-8+1024

    X^2-8+924
    (x- ) (x- )
    this is how far i got. helP!!
    \log(x-16)+\log(x-64)=2

    \Rightarrow \log(x - 16)(x - 64) = 2

    \Rightarrow x^2 - 80x + 1024 = 10^2 = 100

    \Rightarrow x^2 - 80x + 924 = 0

    You could use the quadratic formula here, but this shouldn't be too difficult to factor. We want to find two numbers whose product is 924 and whose sum is -80. Our possible integer factorizations of 924 are (plus or minus) 1\cdot924, 2\cdot462, 3\cdot308, 4\cdot231, 6\cdot154, 7\cdot132, 11\cdot84, 12\cdot77, 14\cdot66, 21\cdot44, 22\cdot42 and 28\cdot33. Of these choices, only -14 and -66 work. So we get

    (x - 14)(x - 66) = 0

    \Rightarrow x = 14\text{ or }x = 66.

    Note, however, that 14 is not a valid solution, since \log(14 - 16) = \log(-2) is not defined over the reals. Thus x = 66 is the only solution.
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