# Plotting to Get a Straight Line

• ialan731
In this particular case, you could use a semilog plot to plot y^2 vs 1/x or ln(y^2) vs 1/x. The slope would be -b and the intercept would be lna.f

## Homework Statement

State what you would plot to get a straight line if experimental (x,y) data are to be
correlated by the following relations, and what the slopes and intercepts would be in
terms of the relation parameters. If you could equally well use two different kinds of plots
(e.g. rectangular or semilog), state what you could plot in each case [the solution to part
(a) is given as an example]

a) y^2=a*e^(-b/x)
solution: construct a semilog plot of y^2 vs 1/x or a plot of ln(y^2) vs 1/x on rectangular coordinates, slope = -b, intercept =lna.

c) 1/ln(y −3) = (1+ a*sqrt( x))/b

e) y = exp(a sqrt(x) +b)

g) y=(ax+b/x)^(-1)

## The Attempt at a Solution

I'm really confused about this. I tried looking in the book, but I couldn't find anything. I feel like the answer should be pretty easy, but I'm just not getting it. If you could help me with one of the questions it should definitely help me solve the other ones. Thanks in advance!

Do you understand how to arrive at that answer you have for (a)?

The answer for a was given. I was looking at it so that I could get an idea as to how to complete the problem, but I don't see how they got it.

y^2=a*e^(-b/x)

Taking the natural log of both sides gives:

log (y^2) = log ( a*e^(-b/x) )

The RHS can be simplified using rules associated with logs, viz., log (x*y) = log(x) + log(y)

= log (a) + ...

Can you fill in the blank?

log(e^(-b/x))

log(e^(-b/x))
This can be simplified. Remember the log we are using is loge

So it would be just log(-b/x)

When you reach an impasse in algebra, it can be enlightening if you substitute numbers and play with the expression. When you think you have found the simplified form, test with numbers to see whether it equals the expression you started out with.

Perhaps first try to simplify log10(10^7)

So it would be just log(-b/x)

You can go further than that.

So it would be just log(-b/x)

$$2\ln{y}=\ln{a}-\frac{b}{x}$$