How to know if there is an inverse or direct relation

[isukatphysics69]

Homework Statement

Homework Equations

log(y) = mlog(x)+log(k)
y=kx^m

The Attempt at a Solution

Determine the exponent m and coefficient k of the power law that best fits your data. Is the acceleration directly or inversely proportional?

Taking some points on the graph to get the slope
(0.78 - 0) / (2.28 - 1.52) = 1.026315789 = slope
finding the y intercept
y = mx + b
b = -mx + y
when y is 0 x is 1.52 so
b = -1.026315789*1.52 = -1.559999999

Now to the question
k = 10^-1.559999999 = 0.02754228
m = 10^1.026315789 = 10.6246783
function is now
0.02754228*x^10.6246783]

This looks like an exponential function. So on the log log graph it looks linear and on the normal graph it looks exponential. I am unsure how to determine if that is inversely or directly proportional. I am thinking inversely.

 

[Tom.G]

Google is your friend on this one, it has over 1 000 000 answers for directly and inversely proportional.
After all, part of learning is learning to find things.

 

[ehild]

Homework Statement

View attachment 225004

Homework Equations

log(y) = mlog(x)+log(k)
y=kx^m

The Attempt at a Solution

Determine the exponent m and coefficient k of the power law that best fits your data. Is the acceleration directly or inversely proportional?

Taking some points on the graph to get the slope
(0.78 - 0) / (2.28 - 1.52) = 1.026315789 = slope
finding the y intercept
y = mx + b
b = -mx + y
when y is 0 x is 1.52 so
b = -1.026315789*1.52 = -1.559999999

Now to the question
k = 10^-1.559999999 = 0.02754228
m = 10^1.026315789 = 10.6246783 wrong.
function is now
0.02754228*x^10.6246783]

This looks like an exponential function. So on the log log graph it looks linear and on the normal graph it looks exponential. I am unsure how to determine if that is inversely or directly proportional. I am thinking inversely.

You denoted the slope by m, and calculated it as m=1.026. The original function is a=k*x^m. This is not an exponential function.

 

[isukatphysics69]

You denoted the slope by m, and calculated it as m=1.026. The original function is a=k*x^m. This is not an exponential function.

Ok i think i see, the slope of the log log graph is actually the power of the "normal" graph

 

[isukatphysics69]

0.02754228*x^1.026315789]
Graphing this give a straight line. So is it correct to say that direct proportionality will be determined by linear growth in both log log graph and "normal" graph

Thinking about this i don't think so, there may be graphs where there are curves but an underlying power law relation.

 

[isukatphysics69]

Wow i am a complete moron. i have figured out this lab report. It has nothing to do with transitioning from log graph to normal graph it is just a matter direct (multiplication) or inverse (division) relationships. i was thinking there is some kind of relation between log log graphs and normal graphs that will determine inverse or direct idk what on Earth i was thinking wow
like once you convert graphs there will be a way to tell a certain relationship idk where i got this idea in my head