Finding the slope of a log-log graph?

[isukatphysics69]

Homework Statement

Homework Equations

The Attempt at a Solution

Wikipedia is saying to use

But when i take the points on my graph i am getting a slope of -0.61 using this formula.
When i use the standard
(y₂ - y₁)/(x₂ - x₁)
i get 1.02 which makes more sense
ok now i have picked some different points and am getting a slope of 2.7
But looking at this graph it looks like the slope is just around 1, so Wikipedia is telling me one thing but the graph looks like the slope is just 1. i have never learned about log log graphs before the prof kind of just threw us into this stuff not sure if i should go with my gut or listen to Wikipedia here. don't want to screw up whole lab report

 

[Orodruin]

The values you have on the axes are the values of $\log(x)$ and $\log(y)$, not those of x and y.

 

[isukatphysics69]

The values you have on the axes are the values of $\log(x)$ and $\log(y)$, not those of x and y.

but if you look at some points like rise = 0.40 run = 1.9
(0.40)/(1.9-1.5) = 1

 

[isukatphysics69]

wait i think i know what you mean I'm not thinking about this proper

 

[isukatphysics69]

So that isn't actually 0.40 it is actually log(0.40)

 

[isukatphysics69]

but i think that would be incorrect because we already logged them when we took the data

 

[Orodruin]

No. What you have is not really a log-log plot in the sense that your axes are linear. You are merely plotting the logs of some functions. What the Wikipedia page is talking about is when the axes are graded logarithmically and you read the values of, in this case, $\Delta m$ and $a$ from them and not the values of $\log(\Delta m)$ and $\log(a)$.

 

[Orodruin]

Could you please limit yourself to one post at a time? A continuous stream of one-line posts makes it very difficult to answer properly and you will benefit from structuring your thinking if you think your posts through more carefully before you submit them.

 

[isukatphysics69]

Yes i am sorry i will do one at a time. Ok i see what youre saying here. So long story short the slope IS actually roughly 1 because we have taken the logs of some values, this isn't a real log log graph

 

[Chestermiller]

The two values of "a" you have shown on your graph are $10^{0.8}=6.31$ and $10^{0.4}=2.51$. The corresponding two values of $\Delta m$ are $10^{2.3}=200$ and $10^{1.9}=79.4$. So, according to their formula,
$$m=\frac{\log(6.31/2.51)}{\log(200/79.4)}=\frac{\log(2.51)}{\log(2.51)}=\frac{0.4}{0.4}=1$$

 

[isukatphysics69]

Ok now i am looking for the power law that best fits data. i am starting with
log(y) = mlog(x)+log(k)
log(y) = log(x)^m+log(k)
10^log(y) = 10^logxm*k
y=x^m*k
now to find the k
10^k
i don't think i should raise 10^m tho to find m tho because it is x^m

 

[isukatphysics69]

i am reading some good information on log log stuff right now i think i might be able to figure out
i have to raise the m to the 10^m , it doesn't make sense if i don't. the slope that i have right now from the log log graph is 1.026315789 so the slope on a normal graph would be 10^1.026315789 = 10.6246783

i am concluding that my data is showing an inversely proportional relation to acceleration and mass