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Math Help - Logarithms & Graphs!

  1. #1
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    Question Logarithms & Graphs!

    Hi people im new to the forums and i have a question to ask

    The question is based on a logarithmic graph which i find difficult, the question states the following:

    Show from the results of voltage and admittance of a circuit in which the law connecting the quantities is of the form V=kY^n . With provided logarithmic paper you need to plot the graph as well as determine the values of K and n. A table of values is provided.

    Voltage (V) 2.88 2.05 1.60 1.22 0.96
    Admittance (Y) 0.52 0.73 0.94 1.23 1.57

    Could someone please help me as i do not understand how to plot this :S
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  2. #2
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    Re: Logarithms & Graphs!

    The point of "logarithmic graphs" is that instead of graphing "V" against "Y" you graph "log(V)" against log(Y). Any equation of the form V= kY^n becomes, after, taking the logarithm of both sides, log(V)= nlog(Y)+ log(k), which has a straight line as graph (of log(V) as a function of log(Y)). And straight lines are easier to recognize than other curves.

    Now if Y= 0.52 and V= 2.88, then log(Y)= -0.2840 and log(V)= 0.4594.
    If Y= 0.73 and V= 2.05, then log(Y)= -0.1367 and log(V)= 0.3118.
    If Y= 0.94 and V= 1.60, then log(Y)= -0.02687 and log(V)= 0.2041.
    If Y= 1.23 and V= 1.22, then log(Y)= 0.2041 and log(V)= 0.001507.
    If Y= 1.57 and V= 0.96 then log(Y)= 0.1959 and log(V)= -0.01773.

    Graph those (log(Y), log(V)) points. I haven't graphed them myself but I suspect that the whole point of this problem is that they lie, a least roughly, along a straight line. A straight line can be written "y= ax+ b". There are a variety of ways of finding "a" and "b"- "a" itself is the "slope" of the line and "b" the "y-intercept" but any two points will give two equations that can be solved for a and b. If you were given a problem like this you are clearly expected to know how to find those coefficients for a straight line.

    Since here, your "x" is "log(Y)" and "y" is "log(V)", "y= ax+ b" becomes "log(V)= alog(Y)+ b". Take 10 (I calculated "common logs" above- if I had used the "natural logarithm I would use "e") to the power of each side, we have 10^{log(V)}= 10^{alog(Y)+ b}= 10^{log(Y^a)}10^{b} or V= (10^b)Y^a.
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  3. #3
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    Re: Logarithms & Graphs!

    Accidental double post.
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  4. #4
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    Re: Logarithms & Graphs!

    Thank you very much for the explanation, much appreciated
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