Is the sum of two exponential function can be equivalent to a third exponential function?

Question

What will be the sum of two exponential functions $2\exp(4 x) + 3 \exp(5 x)$ equivalent to a third exponential function? Is it possible?

Answer 1

Just as goldenration commented, it is impossible to define the sum of two exponential functions as a third exponential function.

However, and this is just a long comment, in some cases, for some specific reasons and at least locally, the sum of two exponential functions can be approximated by a single exponential function and this can be very useful when you want to approximate solutions of equations.

Let us suppose that you want to find many many times a rough estimate of the solution of equation $$2e^{4 x} + 3 e^{5 x}=k$$ $k$ being such that $0\leq x \leq 2$. A preliminary curve fit over that range would show you that over this specific range $$2e^{4 x} + 3 e^{5 x}\approx 3.99132 e^{4.90001 x}$$ So, suppose you want to solve approximately for $k=5000$; using the approximation and logarithms, this will immediately give $x \approx 1.45573$ while the exact solution would be $x \approx 1.45479$.

Answer 2

It is impossible to express $Ae^{ax} + Be^{bx} = Ce^{cx}$ for algebraic real values of $x$ (other than the trivial $x = 0$) and where $A,B,C,a,b,c$ are all rational and nonzero. It's quite easy to prove.

The equation can be expressed as $A(e^x)^a + B(e^x)^b = C(e^x)^c$, which upon substituting $y= e^x$ and rearranging becomes $Ay^a + By^b - Cy^c=0$. Since the coefficients and exponents are all rational and real, this equation has only algebraic solution(s) for $y$. However, $y = e^x$, and it can be proved that $e^x$ is transcendental (and therefore not algebraic) for any nonzero algebraic $x$. This is therefore a contradiction, and no such equation can hold.

However, it might be possible to have irrational solutions for $C$ and $c$. It is likely you're only going to be able to approximate such solutions, as Claude Leibovici has done in his answer. I'm not sure whether all such solutions have to be transcendental, though.

Answer 3

$$ a\mathrm{e}^{bx} = a\sum_{i=0}^{\infty}\frac{(bx)^i}{i!} $$ so if we have $$ a_1\mathrm{e}^{b_1x} + a_2\mathrm{e}^{b_2x} = a_1\sum_{i=0}^{\infty}\frac{(b_1x)^i}{i!} + a_2\sum_{i=0}^{\infty}\frac{(b_2x)^i}{i!} $$ so even if we assume $a_1=a_2$ we have $$ \mathrm{e}^{b_1 x} + \mathrm{e}^{b_2 x} = \sum_{i=0}^{\infty} \frac{(b_1x)^i+(b_2x)^i}{i!} $$ so trying to get it in the form of an exponential function $$ \frac{(b_1x)^i+(b_2x)^i}{i!} = \frac{\left(b_1^i+b_2^i\right)x^i}{i!} $$ so can you make $$ b_1^i+b_2^i = c^i $$ or it looks similar to $$ x^n + y^n = z^n $$ where $z^n >0$ thus for integer values we know it is wrong (Fermat Last Theorem) and in general (I do not know) but it will not hold without special conditions on coefficients for non-integer values.

Answer 4

No, it is not possible. Let $a\ne b$ and suppose $$ Ae^{ax}+Be^{bx}=-Ce^{cx} $$ for all $x$ (the minus sign is irrelevant, it just simplifies the argument).

If we use $x=0$, $x=1$ and $x=2$, we have the linear system in the unknowns $A$, $B$ and $C$: $$ \begin{cases} A+B+C=0 \\ Ae^a+Be^b+Ce^c=0 \\ Ae^{2a}+Be^{2b}+Ce^{2c}=0 \end{cases} $$ whose matrix is $$ \begin{bmatrix} 1 & 1 & 1 \\ e^a & e^b & e^c \\ e^{2a} & e^{2b} & e^{2c} \end{bmatrix} $$ which is a Vandermonde matrix, so its determinant is $$ (e^c-e^a)(e^c-e^b)(e^b-e^a) $$ so the system has just the trivial solution $A=B=C=0$, unless $c=a$ or $c=b$, in which cases we have $B=0$ or $A=0$, respectively.