Question

How would you transform P = Ae^kh into the linear form Y = mX + c?

My attempt:

ln(P) = ln(Ae^kh)
ln(P) = kh ln(A)
Here's where I got stuck, and went
ln(P) = ln(A) + kh
ln(P) = kh + ln(A)

Which doesn't quite seem right..

Answer 1

well you made a mistake in 2nd step .you cant distribute kh to front in this case
you should proceed like

ln(P) = ln(Ae^kh)
ln(P) =ln(A)+ln(e^kh)

Answer 2

now can you do it ?

Answer 3

Thanks! Now:

ln(P) = ln(a) + kh ln(e)
And since ln(e) = 1,
ln(P) = ln(A) + kh

Right?

Answer 4

yep :)

Answer 5

So would that be the final result? Or is there a way to continue until P is on its own?

Answer 6

I think that is enough you have the foem y=mx+c

c=kh
m=1

Answer 7

*form

Answer 8

I understand c = kh, but how did you determine m = 1?

Answer 9

umm...i guess,
ln(P) = ln(A) + kh
y= c+ mx
ln P = y, ln A =c, kh=mX
now m can be taken as k or h, depends on which one is variable, the other one, which is not variable will be taken as slope, m

Answer 10

**now X can be taken as k or h,

Answer 11

actually we need info on what are the constants to determine that

Answer 12

If it is given A is constant you need to consider kh as mx else if kh is constant then you need to consider log A as mx

Answer 13

All the task requires is to prove that P = Ae^kh is valid for a given data set. So to prove this, we have to transform it into Y = mx + b, graph it, determine the y-intercept using the graph and use that as c.

Would considering kh as mx work in this case?

Answer 14

generally, the variable is in the power of 'e'
so, yes.

Answer 15

yep

Answer 16

Sweet, I should be fine from here. Thanks for your help guys, I'll be back if I have any more issues :)