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Math Help - Graphing Logarithms

  1. #1
    Newbie trainman238's Avatar
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    Graphing Logarithms

    Okay, so I know that for graphing logs by hand, you can rewrite it as an exponential... for example, log base 5 of x can be rewritten as 5^? = x...
    But, I was just wanting to be sure... I have log base 3 of (x-5) + 3... so, would the exponential form be 3^? = (x-5) + 3?.... I am pretty sure it is... Because then I have to fill out an x y chart to go with it and then I have to graph it on paper.... So, yeah... just wanting to make sure...
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by trainman238 View Post
    Okay, so I know that for graphing logs by hand, you can rewrite it as an exponential... for example, log base 5 of x can be rewritten as 5^? = x...
    But, I was just wanting to be sure... I have log base 3 of (x-5) + 3... so, would the exponential form be 3^? = (x-5) + 3?.... I am pretty sure it is... Because then I have to fill out an x y chart to go with it and then I have to graph it on paper.... So, yeah... just wanting to make sure...
    Graphing log graphs are no where near that complicated. Is this how you have to do it?

    You can graph any log graph by shifting the general form of the log graph. See below:

    All graphs of the form y = \log_a x looks like that. I do not care what the base a is. They all pass through (1,0), shape like a square root graph and are asymptotic to the y-axis. To get the graph y = \log_3 (x - 5) + 3, you take the graph you see below, shift it 5 units to the right (so the asymptote shifts as well, 5 units, and ends up at the line x = 5). Then shift that graph 3 units up (because of the + 3). Now you would simply solve for the x- and y-intercepts. There are no y-intercepts as you can probably tell. to find the x-intercept, set y = 0 and solve for x. then you will know where you graph touches the x-axis
    Attached Thumbnails Attached Thumbnails Graphing Logarithms-log.jpeg  
    Last edited by Jhevon; April 25th 2008 at 11:35 AM.
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  3. #3
    Newbie trainman238's Avatar
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    thanks man... that REALLY helped a lot..
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  4. #4
    Newbie trainman238's Avatar
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    thanks man... that REALLY helped a lot..
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