Solve equation $\exp(ax)+\exp(bx)=1$

Question

The equation is $$ \exp\left(ax\right)+\exp\left(bx\right)=1, $$ where $a$ and $b$ are known real constants, $x$ is unknown. I would like to have the solution in form of relatively known special function (something like Lambert $W$ function, or generalized hyper-geometric $F$).

Answer 1

If we place $y=\exp(x)$ we can rewrite the problema as: $$y^a+y^b=1$$ Such trinomial equation is a particular case of more general: $$y^a+z y^b=1$$ where we have supposed $a>b$. The solution of the trinomial equation can be found by means of Lagrange inversion series or also by Mellin transform as: $$y(a,b,z)=\frac{1}{a}\sum_{r=0}\frac{\Gamma\left(\frac{1+rb}{a}\right)}{\Gamma\left(\frac{1+rb}{a}+1-r\right)r!}(-1)^rz^r$$ So $$x(a,b)=\log\left[\frac{1}{a}\sum_{r=0}\frac{\Gamma\left(\frac{1+rb}{a}\right)}{\Gamma\left(\frac{1+rb}{a}+1-r\right)r!}(-1)^r\right]$$ References

"The Functions of Mathematical Physics ", 1986, Harry Hochstadt

** EDITED to fix several typos **

Answer 2

There is a new special and multivalued function called Lambert-Tsallis, where you can find details about it at "Physica A: Statistical Mechanics and its Applications 525, 164-170", or to get in contact with a friend and coworker R. V. Ramos.

As mentioned before, your problem can be written like $$y^a + y^b = 1$$.

If you choose $C_1= ln(b/a)$ and $C_2=ln(a)$ and making $C=C_2/C_1$ you will find

$$ y=\left[ C\cdot W_{\frac{1-C}{C}}\left(\frac{1}{C}\right) \right]^\frac{1}{C_1} (1)$$

It must be emphasized that $W_q(x)$ is a multi-valued function and it will generate complex numbers. The manipulations used to arrive at Eq(1) consider that you have to ignore the complex roots and to get only the real roots. Also, the same care with exponentiation! At the end, maybe you want to know the real values of $y$. In this way, you have to calculate all values of $y$ and to get only those reals! It works. Let me give you an example. $$Ex_1: a=4, b=17 $$

it will generate to $W_q(x)$ function the values $$-5.891953651234767 - 0.568776496431273i, -5.891953651234591 + 0.568776496431205i, $$ $$ -2.263956382339662 - 2.956616138268131i, -2.263956382339578 + 2.956616138268151i,$$ $$ 1.184097200785913 - 0.000000000000043i$$

and the final solution $y$ as $$ 1.018598506148727 - 0.243077842499787i, 1.018598506148725 + 0.243077842499787i$$ $$ 0.995769093645394 - 0.172057290937165i, 0.995769093645393 + 0.172057290937163i$$ $$ 0.925272251271377 - 0.000000000000003i$$ where the only correct result is the last one. It happens because you have to ignore since the beginning (for this specific of problem) the complex roots generated by $W_q(x)$

$$Ex_2: a=\pi, b=e^1 .$$ I use these numbers only to support the potencial of $W_q(x)$ function. $ W_q$ result is -0.148963736984781 - 0.000000000000020i and the final solution is $y= 0.788988678082841 - 0.000000000000249i$

Answer 3

Now, getting back to the original problem $$ e^{ax} + e^{bx} =1 \hspace{1cm} (1)$$ If one consider $y=e^x$ at Eq.(1) it could rewritten as $$ y^{a} + y^{b} =1 \hspace{1cm} (2)$$ whose solution was previously given. Thus, the general solution for $x$ will be given by

$$x=\frac{1}{C_1}\cdot ln\left[ C \cdot W_{\frac{1-C}{C}}\left( \frac{1}{C} \right) \right] \hspace{1cm} (3)$$ $Ex_1$ : consider $a=4, b=7.5$, the $W_q(x)$ will produce the result $0.565629288701988 + 0.000000000000003i$ which will give $$ x = -0.124652854199031 + 0.000000000000002i$$

$Ex_2$ : One interesting set of numbers arise when $b=2a$ ($a=3,b=6$, for example). In this case, one has $W_0(1)$ as argument of natural logarithm which generates the roots $$ -1.618033988749864 - 0.000000000000005i, $$ $$ 0.618033988749858 - 0.000000000000032i$$ Indeed, the analytical results correspond to $x= -\left( \frac{1\pm \sqrt{5}}{2} \right)$, it means, $-1$ multiplying the 2 golden numbers that have the property $-a_- =\frac{1}{a_+}$. Thus, this equation truly has infinity analytic solutions given by $$ x_1= \frac{1}{C_1}\cdot \left[ ln(a_+) + i(2k+1)\pi \right]$$ $$ x_2= -\frac{1}{C_1}\cdot ln(a_+) $$ In our example, getting the results generating by Lambert-Tsallis function, $W_q(x)$, we have obtained
$$ 0.160403941686528 - 1.047197551196597i$$ $$ -0.160403941686554 - 0.000000000000017i$$ The imaginary part of the firs solution is $\pi/3$ because for the numbers used $C_1=3$. If one considers to change only this factor for $\pm \pi/3, \pm 3\pi/3, \pm 5\pi/3$ and so on, the number will be solution of Eq. (1) yet

Answer 4

Assuming that "a" and "b" are not zero, this equation does not have any explicit solution except if, say, "b" is a multiple of "a".

So, for the general case, this equation would be solved using a root-finder method such as Newton. What is nice is that, knowing the values of "a" and "b", a reasonably good guess of the solution can be easily made.

If you want to see that working, just give me the "a" and "b" you want and I shall post the path to solution for you.

Answer 5

$$e^{ax}+e^{bx}=1\tag{1}$$ $x\to\ln(y)$: $$y^a+y^b=1\tag{2}$$

For $a,b,x\in\mathbb{N}$, equation (2) is a Diophantine equation.

For rational $a,b$, equation (2) is related to an algebraic equation and we can use the known solution formulas and methods for algebraic equations.

For real or complex $a,b\neq 0,1$, equation (2) is in a form similar to a trinomial equation. A closed-form solution can be obtained using confluent Fox-Wright Function $\ _1\Psi_1$ therefore.

see also: How to isolate $x$ in $a^x + b^x = c$? (For use in medical statistics)

Szabó, P. G.: On the roots of the trinomial equation. Centr. Eur. J. Operat. Res. 18 (2010) (1) 97-104

Belkić, D.: All the trinomial roots, their powers and logarithms from the Lambert series, Bell polynomials and Fox–Wright function: illustration for genome multiplicity in survival of irradiated cells. J. Math. Chem. 57 (2019) 59-106

Answer 6

Suppose without loss of generality, that $a=1$ and $b<1$ and set $y=\mathrm e^{x}$, with $0<y<1$.

• If $b<0$, the minimum of $x\mapsto x+x^b$ for $x\in]0,+\infty[$ is $$(-b)^{\frac1{1-b}}+\left(-b\right)^{\frac b{1-b}}=(-b)^{\frac1{1-b}}+\left(\frac1{-b}\right)^{\frac{-b}{1-b}}>1,$$ because if $b\neq-1$ either $-b$ or $-1/b$ is larger than 1. If $b=-1$ this minimum obviously equals $2$. There is no solution.

• If $b=0$, as $\mathrm e^x+1>1$, there is no solution either.

• If $0<b<1$, the function $y\mapsto 1-y^{b}$ is a decreasing function on $[0,1]$. The equation rewrites $y+y^{b}=1$ or $$ y=1-y^{b}.\tag{1}$$ Thus the solution is of the form $$x(b)=\ln f(b),$$ where $f(b)$ is the fixed point of $y\mapsto 1-y^b$ in $]0,1[$.$$ $$ An algorithm to find the numerical value of $x(b)$: define $z_0\in]0,1[$ and compute the values of $z_n=1-z_{n-1}^b$ for $n\geq 1$. Then the sequence $(z_n)_n$ converges towards $f(b)$.

As it is said in the comments to your questions, even for several values of $b\in\mathbf Q\cap]0,1[$, there is no analytic expression for $x(b)$ or $f(b)$, then there is no possibility for finding a function (hypergeometric series, Lambert, etc...) for all $b\in]0,1[$.