How do you convert a non-linear equation to its linear form?

[Kulla_9289]

How do you convert a non-linear equation to its linear form? For example, if I have [imath]y(a-x)=bx[/imath], how would I convert it to its linear form of [imath]Y=aX+b[/imath]? Do I use substitution as in [imath]ax+b=\frac{bx}{a-x}[/imath]

 

[Dr.Peterson]

How do you convert a non-linear equation to its linear form? For example, if I have [imath]y(a-x)=bx[/imath], how would I convert it to its linear form?

Why do you think a non-linear equation would have a linear form?

Please give an example of what you mean, and also what your ultimate goal is. Why would you want to do this?

 

[Kulla_9289]

Convert [imath]y(a-x)=bx[/imath] to its linear form of [imath]Y=aX+b[/imath]. Do I use substitution as in [imath]ax+b=\frac{bx}{a-x}[/imath]?

 

[JeffM]

Convert [imath]y(a-x)=bx[/imath] to its linear form of [imath]Y=aX+b[/imath]. Do I use substitution as in [imath]ax+b=\frac{bx}{a-x}[/imath]?

A non-linear equation does not have a linear form. And this equation is non-linear.

 

[stapel]

How do you convert a non-linear equation to its linear form? For example, if I have [imath]y(a-x)=bx[/imath], how would I convert it to its linear form of [imath]Y=aX+b[/imath]? Do I use substitution as in [imath]ax+b=\frac{bx}{a-x}[/imath]

Kindly please reply with the full and exact text of the exercise on which you're working, along with a clear listing of your thoughts and efforts so far. This should be helpful in figuring out what's going on. When you reply, please explain where the "substitution" came from, and why you think it applies.

Thank you!

Eliz.

 

[blamocur]

Why do you think a non-linear equation would have a linear form?

Wouldn't that be wonderful ?

 

[Kulla_9289]

The goal is to make this equation a straight line on a graph instead of a curve, etc. For example, if you have a pendulum with length L that moves in time T, you can conduct an experiment and list down the values in a table of the time taken each time you increase the length. Then, when you plot them, it would be a curve, not a straight line. What if you believe T = a√l makes the curve a straight line? You then plot √l against T (square rooting the values of l from the initial experiment) and find the value of a from the straight line.

 

[Kulla_9289]

If you are looking for an application, here it is. The question this on this post is the backbone for the attached problem.

 

[Dr.Peterson]

The goal is to make this equation a straight line on a graph instead of a curve, etc. For example, if you have a pendulum with length L that moves in time T, you can conduct an experiment and list down the values in a table of the time taken each time you increase the length. Then, when you plot them, it would be a curve, not a straight line. What if you believe T = a√l makes the curve a straight line? You then plot √l against T (square rooting the values of l from the initial experiment) and find the value of a from the straight line.

If you are looking for an application, here it is. The question this on this post is the backbone for the attached problem.

I was going to say, it sounds like you're looking for something analogous to log-log or semi-log plotting, transforming the variables; and that's what this new question is about.

But nothing like that will work for your original question,

How do you convert a non-linear equation to its linear form? For example, if I have [imath]y(a-x)=bx[/imath], how would I convert it to its linear form of [imath]Y=aX+b[/imath]? Do I use substitution as in [imath]ax+b=\frac{bx}{a-x}[/imath]

That function has a vertical asymptote, and no monotonic increasing transformation (like the log) can turn that into a line.

 

[Kulla_9289]

The mark scheme says the answer to the original question is x= -b(x/y) + a, Y = x, X = x/y, a = a, b = b

 

[BigBeachBanana]

If you are looking for an application, here it is. The question this on this post is the backbone for the attached problem.

If [imath]\ln y = 0.2[/imath], then what's [imath]y = ?[/imath]
If [imath]\ln y = 0.08[/imath], then what's [imath]y = ?[/imath]

After that, you have 2 coordinate points. Plug them in the general form [imath]y=Ab^x[/imath]. Two equations, two unknowns.

 

[Dr.Peterson]

The mark scheme says the answer to the original question is x= -b(x/y) + a, Y = x, X = x/y, a = a, b = b

Please show us the problem entirely, as given to you! You haven't done that, which makes it very hard to help.

I'm sure the problem doesn't say "convert [this] non-linear equation to its linear form". What does it tell you to do?

 

[Kulla_9289]

This was the instruction: Convert each of these non-linear equations into the form of Y = aX + b, where a and b are constants of each of these questions. State clearly what the variables X and Y and the constants a and b represent. There may be more than a way to do this.

 

[Kulla_9289]

If [imath]\ln y = 0.2[/imath], then what's [imath]y = ?[/imath]
If [imath]\ln y = 0.08[/imath], then what's [imath]y = ?[/imath]

After that, you have 2 coordinate points. Plug them in the general form [imath]y=Ab^x[/imath]. Two equations, two unknowns.

Yes. I got the answer, but the attached file was the preferred method.

 

[topsquark]

This was the instruction: Convert each of these non-linear equations into the form of Y = aX + b, where a and b are constants of each of these questions. State clearly what the variables X and Y and the constants a and b represent. There may be more than a way to do this.

Okay, that is a yet different statement of the question than you have posted before.

Then, yes, [imath]x = -b \dfrac{x}{y} + a[/imath] does satisfy this requirement.

However, if you are trying to plot a data set be aware that your X as you have defined it here, does not behave well at y = 0 and x = a. Personally, I would not call this a "linearized form." But if you had a data set you could find values for b and a from it.

The simplest way to get there is probably to solve for one of the constants, say b:
[imath]y(x - a)= bx[/imath]

[imath]b = \dfrac{y(a - x)}{x} = \dfrac{y}{x} ( x - a)[/imath]

Then after a little thought:
[imath]b \dfrac{x}{y} = a - x[/imath]

and
[imath]x = -b \dfrac{x}{y} + a[/imath]

The procedure is going to be different for each equation you will encounter. Experience is going to be your best instructor.

-Dan

 

[skeeter]

Kepler’s 3rd law of planetary motion, a non-linear relationship …
[imath]T^2 = k \cdot R^3 \text{, where }k = \dfrac{4\pi^2}{GM}[/imath]

[imath]\log{T^2} = \log(k \cdot R^3)[/imath]

[imath]2\log{T} = \log{k} + 3\log{R}[/imath]

[imath]\log{T} = \dfrac{\log{k}}{2}+\dfrac{3}{2} \cdot \log{R^3}[/imath]

analogous to [imath]y = b + mx[/imath]