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Rearranging an equation using log functions

  • Thread starter Rosstickle
  • Start date
  • #1

Homework Statement



Hi guys. I'm trying to rearrage an equation using the natural log and I thought I had it correct but am now struggling. I need to rearange it into the form: y=a0 + a1*C. Where a0 and a1 are some combination of the other variables and numbers below

Homework Equations



H = ((2^(X-1))-1)/(K*(X-1)*C^(X-1))

H,X,K and C are all variables

The Attempt at a Solution



I've tried rearanging using log laws but I now think it might be wrong.


ln(H) = ln(2^(X-1)-1) - ln(K*(X-1)) - (X-1)*ln(C)

But I'm not sure if this is legal considering the equation could be rearranged to:

H = (2^(X-1))/(K*(X-1)*C^(X-1)) - 1/(K*(X-1)*C^(X-1))

Can I then apply log as follows:

ln(H) = ln(2^(X-1))/(K*(X-1)*C^(X-1)) - ln(1/(K*(X-1)*C^(X-1)))

or can I only apply log once to each side, eg:

ln(H) = ln((2^(X-1))/(K*(X-1)*C^(X-1))- 1/(K*(X-1)*C^(X-1)))
 

Answers and Replies

  • #2
BruceW
Homework Helper
3,611
119
ln(H) = ln(2^(X-1)-1) - ln(K*(X-1)) - (X-1)*ln(C)
H = (2^(X-1))/(K*(X-1)*C^(X-1)) - 1/(K*(X-1)*C^(X-1))
These two are fine. But
ln(H) = ln(2^(X-1))/(K*(X-1)*C^(X-1)) - ln(1/(K*(X-1)*C^(X-1)))
Is not right. You can only apply log once to each side (as you said), it would be:
ln(H) = ln((2^(X-1))/(K*(X-1)*C^(X-1))- 1/(K*(X-1)*C^(X-1)))

But What is it you are trying to do? Get C to be the subject of the equation?
 
  • #3
I was trying to rearrange it to the form:

y= a0 +a1*C

or

ln(y) = a0 + a1*ln(C)

I was trying to do this so a program called polymath could determine the a0 and a1 (using experimental date), so that I could find the parameters X and K for the given set of data.

I have since worked it out though. Thank you for your help anyway though.
 

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