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I am curious to know which values of $t \gt 0$ solve the following equation in terms of the constants $a,b,c$.

$a e^{-2 b t} - e^{-2 t} + c e^{-3 b t} + c e ^{- 3 t} = 0$

where

$a \gt 1, b \gt 1, c \gt 0, t \gt 0$.

I would like to know how many roots it has, depending on the values of the constants, and to know the formulas for these roots in terms of the constants, if possible.

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1 Answer 1

up vote 1 down vote accepted

As a start, if $x = e^{-t}$, this becomes $a x^{2b} - x^2 + c x^{3b} + c x^3 = 0$. This can be grouped in various ways.

For example, $a x^{2b} + c x^{3b} = x^2 - c x^3$, or $x^{2b}(a + c x^b) = x^2(1-cx)$ or $x^{2(b-1)}(a + c x^b) = 1-cx$.

At $x = 0$, the LHS is $0$ and the RHS is 1; at $x > 1/c$, the LHS is positive and the RHS is negative, so there is a root between 0 and $1/c$ and none for larger $x$.

Since the LHS is monotone increasing and the RHS is monotone decreasing, there is only one root, and any reasonable method (I would use Newton's) should quickly find it.

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