Questions involving graphs and graphing?

Question

1) We want to graph the function y = x^6 so it looks linear. Should we use a semi-log or log-log graph?
What is the slope and intercept of the straight line? State your reasoning.

2) We want to graph the function y = 5.e^(3.x) so it looks linear. Should we use a semi-log or log-log graph?
What is the slope and intercept of the straight line? State your reasoning.

3) In a clear sentence, what is the usefulness of a log-log graph?

4) In a clear sentence, what is the usefulness of a semi-log graph?

Good Luck! :)

Answer 1

1) You will want to use a log-log plot. That is, you want to plot log y versus log x.

Take the log of your equation:

log y = log (x^6)

this is equal to:

log y = 6 * log x

Now, substitute log y with yP and log x with xP and plot xP on the x-axis and yP on the y-axis. Your result is a line with slope 6:

yP = 6*xP

It will intersect the yP axis at (xP,yP)=(0,0). Note that this is equivalent to saying that it will intersect the (log y) axis at (log x, log y)=(0,0). In other words, the intersection you see on the log-log plot actually represents a LIMITING case.


2) You want to use a semi-log plot. That is, you want to plot log y versus x.

Take the log of the equation:

log y = log[ 5*e^(3x) ]

which is

log y = log 5 + log e^(3x)

which is (I'm assuming log is natural log, but this holds for any log; other bases just scale the result):

log y = log 5 + 3x

Now, as before, substitute yP for log y and you get:

yP = log(5) + 3*x

This is clearly a line that intersects the yP-axis at:

(x,yP) = (0,log(5))

In other words, when x=0, then yP = log y = log(5). Thus, when x=0, y=5.

The slope of this line is 3.


3) A log-log graph makes it easy to characterize power law distributions by their exponent, which will be the slope of the graph.


4) A semi-log graph makes it easy to characterize expoential distributions by their rate, which will be the slope, and their amplitude, which will be their intercept.


Data often comes as either a power law or an exponential. A quick way to test how data is distributed (without having to do some curve fit) is to plot it on a semi-log or a log-log plot. If you get a line on the semi-log, then its exponential. If you get a line on the log-log, then its a power law. As a bonus, you can even read off the important characteristics of each (like exponent, rate, amplitude, etc.).

Answer 2

properly putting it into slope intercept formula is the 1st step we are going to initiate with the 1st one: -4x+y= -4 ...umm..idk if I ought to describe this yet-in basic terms in case- i'm going to you prefer the y by using itself so which you may ought to pass the -4x over to the different part when you consider that its unfavorable you may upload 4x to the two facets and you have got: y= -4 + 4x that's extra of course 4x - 4 you do the comparable for the subsequent one x=a million+(a million/4)y to get the y by using itself: a million)pass the only over by using subtracting a million 2)now you have x - a million= (a million/4)y the y isnt by using itself yet so which you ought to get rid of the (a million/4) by way of fact that's branch you ought to do the alternative-addition so extremely of dividing by using 4 you 3) multiply by using 4 *submit to in concepts that despite you do on one part you ought to do on the different so which you ought to multiply the x AND the -a million by using 4 4)you get 4(x-a million)=y that's 4x - 4=y now you have them the two in slope-intercept sort...and that they are the comparable lol you comprehend the slopes are the numbers in front of the x and the different selection is your y-intercept(the factor the place the line is on the y-axis and x=0) you may desire to have the potential to entice it now y=4x - 4 starts on (0,-4) up 4 over a million and that's it desire it enables...although sorry if it sort of sounds like i'm goin over glaring stuff in basic terms attempting to think of what you may prefer consistent with this sort of question