I'm having trouble proving that there exists only one root for a simple algebra equation

Question

so I have this problem that $$ y = C_1e^{-kx}+C_2e^{-mx} $$

and I need to show that there only exists one root of the equation, not counting the limit at infinity. I tried setting the equation equal to $0$ and solving using logarithm properties, but the $x$'s cancel out that way. Is my algebra wrong, or could someone help push me in the correct direction?

Thank you!

Answer 1

Assuming that $C_1C_2 <0$ and $m \neq k$,

$$C_1\exp(-kx)+C_2\exp(-mx)=0$$ $$C_1\exp(-kx)=-C_2\exp(-mx)$$ $$\exp((m-k)x)=-\frac{C_2}{C_1}$$

then we have $$x = \frac{1}{m-k}\ln \left( -\frac{C_2}{C_1}\right)$$

Answer 2

Set $C_1e^{-kx}+C_2e^{-mx}=0$. Simplify:

$$e^{-kx}(C_1+C_2e^{-(m-k)x})=0$$

Since $e^{-kx}$ is never zero, you're left with $C_1+C_2e^{-(m-k)x}=0$. Since exponential functions are injective, there is at most one point at which $C_2e^{-(m-k)x}=-C_1$. Therefore, the equation has at most one root. Note that it may also have no roots (for example, if both $C_1$ and $C_2$ are positive).