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# Why is the answer to this D?

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1. ???????
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2. Rearranging D, you get . So if you plot log x against log y, it will be a straight line as the equation will have a linear form, , where and .
3. (Original post by RLJCMorrissey)
???????
Because D is the only one that provides a linear relationship when log(y) is plotted against log(x).

...which is in the form as a and b are constants and
4. @B_9710 Hmm, I did think that was the case but surely having logs in there means it doesn't quite come out as a straight line?
5. (Original post by RLJCMorrissey)
@B_9710 Hmm, I did think that was the case but surely having logs in there means it doesn't quite come out as a straight line?
No, it will be a perfect straight line as you are plotting log y against log x. If you had log y =x , by plotting log y against log x you would plot a straight line. That is kind of the point, it's much easier to work with straight lines than curves.
6. (Original post by RLJCMorrissey)
@B_9710 Hmm, I did think that was the case but surely having logs in there means it doesn't quite come out as a straight line?
Why? Anything that can be rearranged to y=mx+c is a straight line. If you can't convince yourself try plotting it
7. I just can't quite get my head around why those logs won't cause it to curve : s
8. (Original post by RLJCMorrissey)
I just can't quite get my head around why those logs won't cause it to curve : s
You are plotting a graph of the form Y = mX + c, with Y on the vertical axis and X on the horizontal axis. The fact that Y and X can be expressed as logarithms of other variables is irrelevant, the point is that they are variable and continuous (well, by continuous in this context I am saying that over a certain interval they take every single real value)
9. (Original post by RLJCMorrissey)
I just can't quite get my head around why those logs won't cause it to curve : s
Well, because the logs are on the axis you're not really plotting them. Try thinking about it in terms of another two variables.
p=log x and q=log y
so you're now plotting
q=log a + bp
Since log a is a constant this is clearly a straight line- it just so happens that q and p are linked to x and y using logs. They can still take any value.
10. (Original post by sindyscape62)
Well, because the logs are on the axis you're not really plotting them. Try thinking about it in terms of another two variables.
p=log x and q=log y
so you're now plotting
q=log a + bp
Since log a is a constant this is clearly a straight line- it just so happens that q and p are linked to x and y using logs. They can still take any value.
Ahhh got it now.

Thanks a lot!

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