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How does using a logarithm help achieve linearity in a dsitribution?Please explain, a...
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(Level 1) Associate Educator, Expert, Newton
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I think this is what your question refers to:
When you have a bunch of data points, and you would like to model them with an equation, one way to do that is to use the "least squares" method to find a straight line that looks like it works.
For example, from the wikipedia page "least squares" comes this picture:
The blue dots are data points, and the red line is a best fit line.
Some relationships aren't linear, however.
The graph is:
(EDIT: hopefully the graph is showing up for you? My computer is only displaying an empty graph??)
Now, suppose we didn't know the equation, but we just had a whole bunch of data points that surrounded the curve `y=5e^(2x)` , and we wanted to figure out that the equation was `y=5e^(2x)`
If you did a least squares analysis on these hypothetical points, you would get a straight line, which wouldn't really match the data.
Try laking the logarithm:
`"log" y = "log" (5 e^(2x)) = "log" 5 + "log" (e^(2x)) = "log" 5 + 2x`
Let's call Y=log (y)
And log (5) is approximately 1.6
So, what you get is:
This is a straight line!
What that means is, if your collection of data points (dots on a graph) looks exponential, then you can take the logarithm of all the y-values, and do least squares on (x, log y) of all the points. You wind up with a straight line (in this example, you will have figured out that your slope is 2 and your y-intercept is about 5). You can convert this back to the exponential relationship by reversing the steps:
`"log" y = 2x+1.6`
`y = e^(2x+1.6) = e^(1.6) e^(2x) = 5e^(2x)`
So you can use a straight line to help you model exponential relationships, by using a logarithm
Or, suppose your data points were all near the curve
But suppose we didn't have the actual equation `y=1.5 x^2.5`
We just had a bunch of dots that were near the curve on the graph. How could we use logarithms to figure out the equation?
`"log" y = "log" 1.5 + "log" (x^2.5) = "log" 1.5 + 2.5 "log" x`
If we write Y = log y, X= log x, we have:
Y = .4 + 2.5 X
Again, this is a straight line. What this means is, if we took looked at the logarithm of the x and y coordinates of all our data points, and plotted those instead, we would get data points that resembled a straight line, and we could do a least squares analysis of it. Working backwards, we could get an equation that modeled the original data, even though the original data weren't in a straight line.
If your data seems to show an exponential relationship, or a power relationship, then you can use logarithms to transform your data into a straight line, use "least squares" to figure out that line (its slope and intercept), and then produce an equation that still models your data.
Posted by mlehuzzah on December 20, 2012 at 7:19 PM (Answer #1)
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