# Logarithms & Graphs!

• Oct 7th 2013, 07:05 AM
Question101
Logarithms & Graphs!
Hi people im new to the forums and i have a question to ask (Happy)

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 1.6 1.22 0.96 Admittance (Y) 0.52 0.73 0.94 1.23 1.57

• Oct 7th 2013, 10:04 AM
HallsofIvy
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 $V= (10^b)Y^a$.
• Oct 7th 2013, 10:13 AM
HallsofIvy
Re: Logarithms & Graphs!
Accidental double post.
• Oct 7th 2013, 10:28 AM
Question101
Re: Logarithms & Graphs!
Thank you very much for the explanation, much appreciated :)