Wikipedia:Reference desk/Archives/Mathematics/2010 April 5

= April 5 =

Determining x in exponentials
Hello! I'm trying to find a way of determining the value of x in an equation of the type:

a*e^(b x)+b*e^(c x)+c*e^(d x)= k

Being a, b, c, d & k constants. Is there anyway for me to compute this? I was looking into logarithms, but maybe there is another way of doing this, considering that I cannot simplify a sum of logarithms...

Thanks for all the help 16:05, 5 April 2010 (UTC)~ —Preceding unsigned comment added by 193.137.208.122 (talk)
 * The solutions to the equation aebx+becx+cedx = k cannot be expressed in closed form, but you may expand the exponentials in taylor series to the second degree and get as an approximation the quadratic equation a(1+bx+b2x2/2)+b(1+cx+c2x2/2)+c(1+dx+d2x2/2) = k, which you can solve explicitly. See also root-finding algorithm. (It is unbelievable that b and c each occurs twice in the equation! ) Bo Jacoby (talk) 16:56, 5 April 2010 (UTC).

Letting u = ex, the equation
 * $$ ae^{bx} + be^{cx} + ce^{dx} = k \, $$

becomes
 * $$ au^b + bu^c + cu^d = k. \, $$

For some values of a, b, c, this can be solved in closed form. For example, if
 * b = 2,
 * c = 1,
 * d = 0,

then the whole thing is a quadratic equation. And if you have 3, 2, 1 instead of 2, 1, 0, then its cubic, but reduces to quadratic if k happens to be 0. Michael Hardy (talk) 20:18, 5 April 2010 (UTC)