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### Equation of Straight Line on the Log-Log Scale

```Date: 03/06/2006 at 00:41:55
From: hard stone
Subject: straight line equation on the log-log scale

I have a log-log graph with a straight line on it, and I want to find
the line's equation.  The x-axis is scaled as 0.01, 0.1, 1, 10, 100
and the y-axis is 10, 100, 1000, 10000.

I am using two points on the given line, (100,1) and (600,0.2) and
finding the slope m = -1 and the y-intercept b = 101 so my equation is
y = -1x + 101.

But none of the points on the given line check when I put them in that
equation except (100,1), so that can't be the right equation.  What am
I doing wrong?

```

```
Date: 03/06/2006 at 09:51:17
From: Doctor Peterson
Subject: Re: straight line equation on the log-log scale

Hi, Hard Stone.

A straight line on the log-log scale is not given by

y = mx + b

but by

log(y) = m log(x) + b

since the actual locations on the graph are the logs of the
coordinates x and y.

This in turn means the same as

y = 10^b x^m

(by raising 10 to the power on each side).

So in order to find the parameters of this equation, which I'll
simplify to

y = c x^m

you want to plug x and y for two points into either form of the
equation and solve for the parameters.  In this case, you can solve

1 = c 100^m
0.2 = c 600^m

You might start by dividing one equation by the other to eliminate
c, and then take a log.

Alternatively, you can solve

log(1) = m log(100) + b
log(0.2) = m log(600) + b

which might be more familiar.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Linear Equations
High School Logs

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