1. ## simplify logarithm

hi
i have the following
ln t = B1 + B2.T + B3.X + B4.X^2 + B5.X^3
where B1,B2,B3,B4,B5 are constants and X = ln y
i need this in a form t = C1 (y^C2) (e^(-C3/T))
where C1,C2,C3 are constants.
ln and e are natural log terms.
can this be done?
thanks
jas
p.s. i know e^(B3.X) = e^(B3. ln y) =e^(ln y^B3) = y^B3
but i'm not sure how to get the other terms B4.X^2 = B4 (ln y) (ln y)
and B5.X^3 = B5 (ln y) (ln y) (ln y)
in a similar form of y^??? (or if its possible!!)

2. Originally Posted by jas
hi
i have the following
ln t = B1 + B2.T + B3.X + B4.X^2 + B5.X^3
where B1,B2,B3,B4,B5 are constants and X = ln y
i need this in a form t = C1 (y^C2) (e^(-C3/T))
where C1,C2,C3 are constants.
ln and e are natural log terms.
can this be done?
thanks
jas
p.s. i know e^(B3.X) = e^(B3. ln y) =e^(ln y^B3) = y^B3
but i'm not sure how to get the other terms B4.X^2 = B4 (ln y) (ln y)
and B5.X^3 = B5 (ln y) (ln y) (ln y)
in a similar form of y^??? (or if its possible!!)
Taking the exponential of both sides,
$\displaystyle t= e^{B1+ B2T+ B3X+ B4X^2+ B5X^3}= e^{B1}(e^{T})^B2(e^x)^{B3}(e^X)^{2B4}(e^X)^{3B5}$
Since x= ln y, $\displaystyle e^X= e^{ln y}= y$ so that is the same as $\displaystyle t= e^{B1}e^{B2T}y^{B3}y^{2B4}y^{3B5}$$\displaystyle = e^{B1}e^{B2T}y^{B3+ 2B4+ 3B5}$.

So taking $\displaystyle C1= e^{B1}$, $\displaystyle C2= B3+ 2B4+ 3B5$, [tex], and $\displaystyle C3= -B3$ you will have $\displaystyle t= C1(y^{C2})e^{-C3T}$. I do not believe you can choose the constants so that it is $\displaystyle e^{-C/T}$ rather than $\displaystyle e^{-CT}$.

3. by
ln t = B1 + B2.T + B3.X + B4.X^2 + B5.X^3
i mean
ln t = B1 + B2.T + B3.(ln y) + B4.(ln y)(ln y) + B3.(ln y)(ln y)(ln y)
so (ln y)(ln y) is not the same as (ln y^2) = 2 ln y ?

so i don't know if the simplification you have is true?