Split ln() of two exponential summands

1. Jul 14, 2014

globi

1. The problem statement, all variables and given/known data
Dear all

I am calibrating a temperature measurement model and I am stuck with an equation. The variable z is given; x and y represent two regression terms with common regressors - which I will solve for a specific regressor in a second step.

2. Relevant equations

ln(e^x) = ln(z) + ln((e^x) + (e^y))

I need all x and y non-exponential, so that I can solve the equation for x and y. Now I am confused with the properties of ln((e^x) + (e^y)), this is a bracket of two exponential summands with two different exponents.

3. The attempt at a solution
As far as I know, the term ln(a+b) can not be rewritten that easily. But I was wondering if this changes if both summands are exponential, as in my case.

How can I solve the equation for x and y?
Thank you.

2. Jul 14, 2014

HallsofIvy

Staff Emeritus
I presume you know that ln(a)+ ln(b)= ln(ab) so your equation is $$ln(e^x)= ln(z(e^x+ e^y). Now take the exponential of both sides. I also presume you know that you cannot solve a single equation for both x and y. It is relatively easy to solve for z or y. To solve for x I think you would need the "Lambert W function". 3. Jul 14, 2014 Ray Vickson If you know $y$ and $z$ the solution is easy: [tex] e^x = \frac{z \, e^y}{1-z}$$

4. Jul 15, 2014

globi

Thank you for your answers. Maybe I was not clear. But as I wrote, only z is known a priori. x and y both represent two regression terms with common regressors, i.e. they have the same variables (which are known) and partially the same coefficients. In the end, I am interested in those coefficients (the relation amongst all coefficients is known, so that I can reduce them to one coefficient).
This is why I need to rewrite the equation in a way that I have both x and y in a non-exponential form. I will then fill in the regression terms for x and y and then solve the whole equation for the coefficients of interest.

HallsofIvy, your suggestion goes actually back in the direction I started. The original tool calibration equation is e^x= z*(e^x+ e^y) with x and y representing regression terms. Then I introduced ln in order to "get x and y down from e" and applied the rule you mentioned and ended with ln(e^x) = ln(z) + ln((e^x) + (e^y)). The left side is clear: I take the ln and get x; ln(z) is also fine (as z is known); but what is with ln((e^x) + (e^y)) ?

I hope now my problem is clear and any help is appreciated.

5. Jul 15, 2014

Ray Vickson

Your problem is very unclear, so let me describe what I THINK you might be saying. You have some independent variables, say u and v and two dependent variables x and y, for which you propose some formulas x = f(u,v) and y = g(u,v), perhaps of the simple forms x = a + b*u + c*v and y = k + m*u + n*v. You want to determine the coefficients a,b,c,k,m,n in such a way that for a given value z we have
$$e^{a + bu + cv} = z \left( e^{a + bu + cv} + e^{k + mu + nv} \right)$$
for all $u,v$ (or at least for some $u,v$) and also that the functions $f(u,v) = a + bu + cv, g(u,v) = k + mu + nv$ satisfy some kind of least-squares criterion (that is, you have a constrained least-squares problem--- constrained by the equation written above). Does that describe your problem, at least roughly? If not, you will need to present some details (such as the forms of x and y, the nature of the variables, etc.) and provide a more extensive explanation.