Finding Proportional Relationships using Log-Log Graphs

[Stormblessed]

Homework Statement

[/B]
Given a a log-log graph with 8 lines, you must determine the equation of each line in its original relationship. The slope of the graph (m) gives the power of the original relationship.

Examples:

if m = 2, 3; then y ∝ x^2, x^3, etc.
if m = -1, -2; then y ∝ 1/x, 1/x^2, etc.
if m = 1/2, 1/3; then y ∝ √x, ∛x, etc.
if m = 2/3; then y ∝ ∛x^2The antilog of the y-intercept (b) of the line gives the proportionality constant (or magnitude of the slope) of the original relationship.

Note: Worksheet is uploaded

Homework Equations

[/B]
y = (antilog b) (X^m) ← To find proportionality constant

m = Δ Log y/ Δ Log x ← To find slope of log-log line

The Attempt at a Solution

I was able to determine the equations of the original relationships for lines #1-4, being:

Line #1: y = 100x
Line #2: y = 100x^2
Line #3: y = x^3
Line #4: y = 10(√x)

However, I am completely stumped on determining the equations for lines 5-8, as the slopes that I calculated are not as easy to convert as the first four lines. So, I found the slopes for lines 5-8, which are:

Line #5: m = -0.9
Line #6: m = -1.5
Line #7: m = -3.8 ≈ -4
Line #8: m = 4/3

I still do not know how to convert these values into the X^m values, as was done for lines 1-4, because these numbers are a little bit wonky.

I also used the antilog for the y-intercept of lines 5-8 to find the magnitude of the proportionality constant:

Line #5: y = 3.2 x 10^5
Line #6: y = 1.0 x 10^5
Line #7: y = 3.2 x 10^4
Line #8: unable to find y-intercept, so I could not determine value of proportionality constant

How do I turn the slopes of the log-log lines into the powers for the original relationships and use that to find the equation of the original line? Also, how do I find the y-intercept of Line #8? An explanation of this would really be helpful.

Note: my understanding in math is at a grade 11 level ( I don't know calculus).

Thanks

 

[Chestermiller]

Suppose these were ordinary x-y plots rather than log-log plots. Would you be able to determine the equation for each of the straight lines? If so, what is your x-y equation for line 5? (Your present determination of the slope is not correct). What would be your x-y equation for line 8?

 

[Stormblessed]

The x-y equation that I got for Line #5 is y = -0.9 x + 5.48. But I still do not know how to convert this into the original relation form.

 

[Chestermiller]

The x-y equation that I got for Line #5 is y = -0.9 x + 5.48. But I still do not know how to convert this into the original relation form.

Now you write $\log{Y}=5.48-0.9 \log{X}=5.48+\log{X^{-0.9}}=\log{(3.02\times 10^5)}+\log{X^{-0.9}}=\log{(3.02\times 10^5X^{-0.9})}$
So, $$Y=3.02\times 10^5X^{-0.9}$$

For line 8, if y = mx + b, the x intercept (y = 0) is $x=-b/m$.