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Math Help - One more question

  1. #1
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    One more question

    Just a couple more im confused about


    I have one question that asks me to 'apply the laws of logarithms' to change the form of the following equation

    One more question-equations.png

    Once again the samples are way easier and dont prepare me I need to change the form so that I can describe the transformations
    the result I got was

    3log_3(x-4)-1

    It feels as if I'm doing it wrong though could you please show me how to get the final form so I can do the transformations?
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  2. #2
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    Re: One more question

    Quote Originally Posted by Rayzala View Post
    Click image for larger version. 

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    Once again the samples are way easier and dont prepare me I need to change the form so that I can describe the transformations
    the result I got was 3log_3(x-4)-1
    I get $3\log_3(x-4)-7$ Recall $3\log_3(9)=6$
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  3. #3
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    Re: One more question

    Quote Originally Posted by Plato View Post
    I get $3\log_3(x-4)-7$ Recall $3\log_3(9)=6$
    So to find the transformations I would use the first function you gave me right?

    Would the description of the transformation be the following?:

    The base function is y=log_3x, stretched vertically by 4, and translated right by 4 and down by 13.
    Last edited by Rayzala; March 3rd 2014 at 05:26 PM.
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  4. #4
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    Re: One more question

    Quote Originally Posted by Rayzala View Post
    So to find the transformations I would use the first function you gave me right?
    Would the description of the transformation be the following?:
    The base function is y=log_3x, stretched vertically by 4, and translated right by 4 and down by 13.
    OH! come on! Learn the basics.

    $\dfrac{1}{5}\log_3(9x-36)^{15}-13\\3\left[\log_3(9(x-4))\right]-13\\3\left[2+\log_3(x-4)\right]-13\\6+3\left[\log_3(x-4)\right]-13$
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  5. #5
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    Re: One more question

    Stretched vertically by 5, and translated right by 4 and down by 13?
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