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Thread: logarithm with a < 1

  1. #1
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    logarithm with a < 1

    Hi;
    I have the function f(x) = -1/2log(3x - 3) + 5.

    I have a table of parent function values x:1/9,1/3,1,3,9 and y:-2,-1,0,1,2.

    I'm ok with all transformations eccept when I have an a value less than 1.
    I know its vertical compression by a factor of a but how do I apply it to my y values to get new values?

    Thanks.
    Last edited by anthonye; Jan 23rd 2015 at 11:49 AM.
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  2. #2
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    Re: logarithm with a < 1

    Quote Originally Posted by anthonye View Post
    Hi;
    I have the function f(x) = -1/2log(3x - 3) + 5.

    I have a table of parent function values x y
    1/9 -2
    1/3 -2
    1 0
    3 1
    9 2

    I'm ok with all transformations eccept when I have an a value less than 1.
    I know its vertical compression by a factor of a but how do I apply it to my y values to get new values?

    Thanks.
    I need some clarification. In your table are you stating that x = 1/9 gives you a y value of -2? If so log(3 (1/9) - 3) = log(1/3 - 3) = log(-8/3) which is not a real number.

    -Dan
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  3. #3
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    Re: logarithm with a < 1

    Oh I think I see, If I have an x value<1 and a<1 do I always end up taking the log of a negative?
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  4. #4
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    Re: logarithm with a < 1

    Quote Originally Posted by anthonye View Post
    Oh I think I see, If I have an x value<1 and a<1 do I always end up taking the log of a negative?
    What's a?

    The argument of the log has to be a positive number. So your condition is 3x - 3 > 0. So the only permitted x value in your list are 3 and 9. There is a problem with your y values though. For example:
    y(3) = -\frac{1}{2} log(3 \cdot 3 - 3) + 5= -\frac{1}{2} log(6) + 5 \neq 1

    What is your parent function?

    -Dan
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  5. #5
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    Re: logarithm with a < 1

    Quote Originally Posted by anthonye View Post
    Hi;
    I have the function f(x) = -1/2log(3x - 3) + 5.

    I have a table of parent function values x:1/9,1/3,1,3,9 and y:-2,-1,0,1,2.

    I'm ok with all transformations eccept when I have an a value less than 1.
    I know its vertical compression by a factor of a but how do I apply it to my y values to get new values?

    Thanks.
    given the table values, it looks as though your parent function is y = \log_3{x} , a base 3 log.

    now, if f(x) is also a base 3 log function (I'm just guessing here ... )

    f(x) = -\frac{1}{2} \log_3(3x-3) + 5

    f(x) = -\frac{1}{2} \log_3[3(x-1)] + )5

    f(x) = -\frac{1}{2} \left[\log_3{3} + \log_3(x-1) \right] + 5

    f(x) = -\frac{1}{2} \left[1 + \log_3(x-1) \right] + 5

    f(x) = -\frac{1}{2}  - \frac{1}{2} \log_3(x-1) + 5

    f(x) = -\frac{1}{2} \log_3(x-1) + \frac{9}{2}

    best I can make out, it looks like f(x) is the parent function transformed as follows ...

    a horizontal shift right one unit, a vertical compression by a factor of 1/2, a reflection over the x-axis, and finally a vertical shift up 4.5 units

    note the transformed function has domain x > 1
    Last edited by skeeter; Jan 23rd 2015 at 04:13 PM.
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  6. #6
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    Re: logarithm with a < 1

    I have a table of parent function values x:1/9,1/3,1,3,9 and y:-2,-1,0,1,2.
    transformed x-values (all shifted right one unit) ...

    10/9 , 4/3 , 2 , 4 , 10

    transformed y-values (after the stated compression, reflection, and vertical shift) ...

    5.5 , 5 , 4.5 , 4 , 3.5
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  7. #7
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    Re: logarithm with a < 1

    Yes parent function f(x) = log3(x) '' that 3 is the base''

    skeeter that still gives me a negative in the log so how did you get the values?

    Thanks.
    Last edited by anthonye; Jan 24th 2015 at 02:27 AM.
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  8. #8
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    Re: logarithm with a < 1

    All x-values shift to the right one unit ... look at post #6
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  9. #9
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    Re: logarithm with a < 1

    Yes just realized and log on.

    Thanks again.
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