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Math Help - graphing logs

  1. #1
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    graphing logs

    could someone explain to me how to graph these?

    #1. y = 4log2x (the 2 is the base)
    #2. y = log0.2x^2 (the 0.2 is the base)

    thank you.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by checkmarks View Post
    could someone explain to me how to graph these?

    #1. y = 4log2x (the 2 is the base)
    #2. y = log0.2x^2 (the 0.2 is the base)

    thank you.
    the fail safe way to draw any graph is to find:
    - intercepts
    -asymptotes
    -end behaviors

    and if you do caclulus
    -critical points
    -concavities

    but these are log graphs, all log graphs have the same shape (the different bases only dictate how steep the graph is). in the general case, the log graph is asymptotic to the y-axis. it goes from negative infinity on the left, passes through (1,0) and goes to positive infinity. this is always the case unless the graph is shifted horizontally or vertically. neither of these graphs are shifted in such ways, so their graphs will look like the one below


    This is my 2th post!!!!!!!!!!!!!
    Attached Thumbnails Attached Thumbnails graphing logs-log.jpg  
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  3. #3
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    Quote Originally Posted by checkmarks View Post
    could someone explain to me how to graph these?

    #1. y = 4log2x (the 2 is the base)
    #2. y = log0.2x^2 (the 0.2 is the base)

    thank you.
    Hello,

    as Jhevon pointed out all log-graphs have the same shape (more or less). If you know the graph of y=\ln(x) then you get the graphs of other log-function by a vertical dilation (to the x-axis) with a constant scale-factor (not certain if I used the correct expressions here).

    to #1: From
    y=4 \cdot \log_2(x)\ \Longleftrightarrow \ y=4 \cdot \frac{\ln(x)}{\ln(2)}=\frac{4}{\ln(2)} \cdot \ln(x) \approx 5.77 \cdot \ln(x)

    As you can see the y-values of y=ln(x) are multiplicated by the constant factor 5.77

    to #2:From
    y= \log_{\frac{1}{5}}(x^2)\ \Longleftrightarrow \ y=\frac{\ln(x^2)}{\ln \left(\frac{1}{5} \right)}=\frac{2}{\ln \left(\frac{1}{5} \right)} \cdot \ln(|x|) \approx -1.243 \cdot \ln(|x|)
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