How can i show that if there is a linear relationship between logX and logY then there is a Power Relationship between X and Y?
Thanks
Since there is a linear relationship, you know that $\displaystyle \log {(y)} = a \times \log {(x)}$ (I'll spare you the y-intercept). This is equivalent to $\displaystyle \log{(y)} = \log{(x^a)}$ (see logarithm laws). Exponentiating both sides yields :
$\displaystyle y = x^a$
Which is clearly a power relationship. I've left something for you to work out : how would you proceed if the linear relationship between the logarithms had a y-intercept different from zero ? (although you clearly mentioned linear which implicitly means there is no y-intercept, that case being we call the relationship affine)
Well if your relationship is in the form $\displaystyle \log{(y)} = a \log{(x)} + b$, then yes you are going to need it. Here is what I would do :
$\displaystyle \log{(y)} = a \log{(x)} + b$
Say $\displaystyle b = \log{(\beta)}$. The equation becomes :
$\displaystyle \log{(y)} = a \log{(x)} + \log{(\beta)}$
Use the laws of logarithms :
$\displaystyle \log{(y)} = \log{(x^a)} + \log{(\beta)}$
Use them again :
$\displaystyle \log{(y)} = \log{(\beta x^a)}$
And then, and only then, exponentiate both sides :
$\displaystyle y = \beta x^a$
Now substitute back $\displaystyle b$ instead of $\displaystyle \beta$ : $\displaystyle \beta = n^b$, where $\displaystyle n$ is the base of your logarithms (if not stated, assume it is ten). You get :
$\displaystyle y = n^b x^a$
Which is still a power relationship. Does it make sense ?
So would:
$\displaystyle
y = n^b x^a
$
be the same as
$\displaystyle
y = ax^n
$
Uhh there's different variables...
But my data is modelled by the power function: $\displaystyle y=ax^n$
And I need to use all this to find the values of A and N using my x and y values given.
Which are:
X1: 0.241 Y1: 0.387
x2: 1.881 Y2: 1.542
X3: 29.457 Y3: 9.539
Yes, it's different variables, don't be scared, sometimes this gets messed up because of awkward variable choice and you have to put it back in order. It's equivalent
So if I get it, you are given :
$\displaystyle y = a x^n$
And you want to find $\displaystyle a$ and $\displaystyle n$ from various pairs of values $\displaystyle (x,y)$, is that right ?
Do you have to linearize your data, or can you use algebraïc methods ? Because there is actually one pretty neat and quick technic you can use there
It says to:
Investigate and evaluate the validity of your results (which I obtained a different way) by analysing this rule (The one we're discussing).
Hence find the values of A and N and comment on the strengths and limitations of the model found in part c.
I've got everything down pat, except for the investigating, evaluationg and finding the values.
I'd assume anyway of finding it would be fine as it hasn't been noted to use a specific way
Alright then, let's use the algebraïc method to extract $\displaystyle a$ and $\displaystyle n$ from data
Your data is modelled by the formula $\displaystyle y = ax^n$.
You need at least two pairs $\displaystyle (x,y)$. You got them :
$\displaystyle (0.241, 0.387)$
$\displaystyle (1.881, 1.542)$
Let us denote this data $\displaystyle (x_1 , y_1)$, $\displaystyle (x_2 , y_2)$. You have :
$\displaystyle y_1 = a x_1^n$
$\displaystyle y_2 = a x_2^n$
This is where we get smart ! We can divide them together
$\displaystyle \frac{y_1}{y_2} = \frac{a x_1^n}{a x_2^n}$
This is the same as :
$\displaystyle \frac{y_1}{y_2} = \frac{x_1^n}{x_2^n}$
Nice ! We only have one unknown left ! We can solve for $\displaystyle n$ :
$\displaystyle n = \frac{\log{(y_1 y_2)}}{\log{(x_1)}+\log{(x_2)}}$
Use your data table to substitute the correct $\displaystyle x_1$, $\displaystyle y_2$, ... values.
Now that you know $\displaystyle n$, use it to easily solve for $\displaystyle a$ in one of the equations I first came up with. You will have found $\displaystyle n$, and $\displaystyle a$
Does it make sense ? The key competency to grasp here was the division of both equations to get rid of the unknown $\displaystyle a$ in order to find $\displaystyle n$, otherwise we would have two unknowns INVOLVING POWERS and that would be longer to solve.
Ah. I will explain this a little better. Forget every other variable in your problem. This is a general theorem.
Say you have an equation in the form :
$\displaystyle \log{(y)} = m \log{(x)} + c$
In order to exponentiate properly without leaving any artifacts, we need to have everything in logs. It is easy to put the $\displaystyle m$ inside :
$\displaystyle \log{(y)} = \log{(x^m)} + c$
(From laws of logarithms)
What about the $\displaystyle c$ ? We're not going to be able to do anything, since it's addition ! But !
Say you define a new letter, say $\displaystyle k$, such as $\displaystyle \log{(k)} = c$. We can substitute this letter into our equation :
$\displaystyle \log{(y)} = \log{(x^m)} + \log{(k)}$
Now we can combine both logs on the right side (laws of logarithms) :
$\displaystyle \log{(y)} = \log{(k x^m)}$
Good ! Let's exponentiate both sides ! We get !
$\displaystyle y = k x^m$
All right. Now, $\displaystyle k$ makes no sense for us, we only used it to allow exponentiation on both sides. We want $\displaystyle c$, not $\displaystyle k$. What now ? Well, we know that $\displaystyle \log{(k)} = c$ (this is how we chose to define $\displaystyle k$), so this is equivalent to saying that $\displaystyle k = 10^c$ (assuming that the logarithms are in base ten, if not just change the base). So we substitute back :
$\displaystyle y = 10^c x^m$
Maybe putting brackets will help you visualize better :
$\displaystyle y = (10^c) x^m$
And this is strictly equivalent to $\displaystyle \log{(y)} = m \log{(x)} + c$.
Does it make sense ?
No problem, that's what MHF is all about !
Thanks ^^' But if you drop in from time to time, you might be able to see pure geniuses here, people far, far, far, better than me. So you see, you can ask about any question
PS : the forum uses a special concept of "thanked" button that allow the original poster to select the answer (or answers) that actually answers his question so that other people can find it quickly and so that credit is given to the actual answerer. If you mind ... (nah just joking, it's not an obligation or anything, it's just recommended)