1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Split ln() of two exponential summands

  1. Jul 14, 2014 #1
    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. jcsd
  3. Jul 14, 2014 #2


    User Avatar
    Science Advisor

    I presume you know that ln(a)+ ln(b)= ln(ab) so your equation is [tex]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".
  4. Jul 14, 2014 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    If you know ##y## and ##z## the solution is easy:
    [tex] e^x = \frac{z \, e^y}{1-z} [/tex]
  5. Jul 15, 2014 #4
    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.
  6. Jul 15, 2014 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    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
    [tex]e^{a + bu + cv} = z \left( e^{a + bu + cv} + e^{k + mu + nv} \right) [/tex]
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted