Showing Linear Independence of $\, \{e^x, \, e^{-x}\} $ We examine the equation

$$ae^x + be^{-x} = 0$$

My book say manipulate to get

$$ae^{2x} = -b$$

and "The two members are identical for all $x$ only if 

$$a =b = 0 \, \text{ "}$$

[Therefore the set in the title is L.I.] 



My confusion is that I can easily choose $a = 2, b = -3$ and find an $x$ where the equation is satisfied. I can repeat this process for every $x \in \mathbb{R}$. 

The definition my book uses for L.I. is 


  A set of functions  \begin{equation}  \{y_1, y_2, \dots, y_n\}
 \end{equation} defined on $(a,b)$ is linearly independent if the
   linear combination for the set  \begin{equation} c_1y_1+c_2y_2 +
 \cdots + c_ny_n = 0 \end{equation} implies that for all $x$ the only solution is 
   \begin{equation} c_1= c_2=\cdots =c_n = 0 \end{equation}


I'm guessing that means the $c_i$ must be fixed, and that's where I'm wrong. I don't see anything in the definition saying that $c_i$ must be fixed, nor anywhere I look; is it implied? The nature of the definition hints that there might be more than one solution depending on the function set (making the set L.D.), which to me means that the set of constants does not have to be fixed.
The $c_i$ do have to be fixed: since we're talking about linear independence of the set of functions, we're looking for a linear combination of functions that add to the zero function, i.e., the function which returns zero for all $x$. This function is fixed by fixed constants $c_i$. I agree that the definition you posted is a bit unclear on this point.
Take natural logarithm from both sides of equality $ae^x=-be^{-x}$, then you can see that $x$ should be $\frac{ln(a)+ln(-b)}{2}$.