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Math Help - Logarith question.

  1. #1
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    I have to solve:

    (log(x^2) + log(x^4))/log(90x) = 3

    I've been on it for 25 minutes and have gotten as far as:

    log(x)/log(90x) = 1/2

    And I'm not sure if I'm even going the right path lol, just trying to play around with properties of logs and not sure how to maneuver this. Thanks for any help
    Last edited by mr fantastic; September 20th 2010 at 06:55 PM. Reason: Moved from another thread.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by DannyMath View Post
    I have to solve:

    (log(x^2) + log(x^4))/log(90x) = 3

    I've been on it for 25 minutes and have gotten as far as:

    log(x)/log(90x) = 1/2

    And I'm not sure if I'm even going the right path lol, just trying to play around with properties of logs and not sure how to maneuver this. Thanks for any help
    Applying the laws of logarithms to reduce the left hand sid to an exoression in \log(x) :

    \dfrac{\log(x^2)+\log(x^4)}{\log(90x)}=\dfrac{2\lo  g(x)+4\log(x)}{\log(90)+\log(x)}

    and you should be able to finish from there.

    CB
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  3. #3
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    Hello, DannyMath!

    What you did was correct . . .


    \text{Solve for }x\!:\;\;\dfrac{\log(x^2) + \log(x^4)}{\log(90x)} \:=\: 3

    Note that: .  x \ne 0


    We have: . \log(x^2\cdot x^4) \;=\;3\log(90x) \quad\Rightarrow\quad \log(x^6) \:=\:3\log(90x)

    . . . . . . . . . . 6\log(x) \:=\:3\log(90x) \quad\Rightarrow\quad 2\log(x) \:=\:\log(90x)

    . . . . . . . . . . . \log(x^2) \:=\:\log(90x) \quad\;\;\Rightarrow\qquad \quad\; x^2 \:=\:90x


    Then: . x^2 - 90x \:=\:0 \quad\Rightarrow\quad x(x-90) \:=\:0


    Therefore: . \rlap{//////}x \,=\,0,\;\;x\,=\,90
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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, DannyMath!

    What you did was correct . . .



    Note that: .  x \ne 0


    We have: . \log(x^2\cdot x^4) \;=\;3\log(90x) \quad\Rightarrow\quad \log(x^6) \:=\:3\log(90x)

    . . . . . . . . . . 6\log(x) \:=\:3\log(90x) \quad\Rightarrow\quad 2\log(x) \:=\:\log(90x)

    . . . . . . . . . . . \log(x^2) \:=\:\log(90x) \quad\;\;\Rightarrow\qquad \quad\; x^2 \:=\:90x


    Then: . x^2 - 90x \:=\:0 \quad\Rightarrow\quad x(x-90) \:=\:0


    Therefore: . \rlap{//////}x \,=\,0,\;\;x\,=\,90
    UGH I was getting so frustrated and you've made it so clear. Thank you! *sigh of relief*
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by DannyMath View Post
    UGH I was getting so frustrated and you've made it so clear. Thank you! *sigh of relief*
    I'm afraid he has actually made rather a meal of it.

    Which I suppose is OK if you are going to spoon feed someone.

    CB
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