Showing linear independence of $\{5, e^{ax}, e^{bx}\}$ The question is as above. 
I have not learnt about the Wronskian or the likes.
I suppose that I'll have to do something with differentiation or trying different x values.
EDIT: 


*

*$a$ and $b$ are non-zero and not equal to each other. I should have mentioned this earlier. 

*I have not learnt about the Wronskian formally and doubt that I should use it. It is nice to learn new methods and new theorems, however.
 A: Let they be linearly dependent, i.e.
$$
A\cdot5+Be^{ax}+Ce^{bx}\equiv0
$$
for $A$, $B$, $C$ not all equal to zero. Then it is true for example for $x=0$, 1, 2, i.e.
$$
5A+B+C=0, 5A+Be^a+Ce^b=0, 5A+Be^{2a}+Ce^{2b}=0.
$$
The determinant is equal to
$$
\begin{vmatrix}
5 & 1 &1\\
5 & e^a &e^b\\
5 & e^{2a} & e^{2b}
\end{vmatrix}.
$$
One can verify (please compute) that it in nonzero. But it means, that there is only one solution $A=B=C=0$. A contradiction.
BTW: Methods with wronskian are easier to use.
A: (I'm assuming $a\ne b$, since otherwise they are dependent.)
One thing you might notice is that these all grow at different rates. A way to use that is the following:
If they are linearly dependent, we would have:
$$A\cdot5+Be^{at}+Ce^{bt}=0$$
where the coefficients are nonzero. 
WLOG, assume $a<b$. (If $a>b$, just switch around $a$ and $b$ in the following argument.) I'll also assume that they are both positive, but it's not hard to modify the argument for if they're negative.
Divide everything by $e^{bt}$ (we can do this since it's never equal to zero):
$$5Ae^{-bt}+Be^{(a-b)t}+C=0$$
Taking the limit as $t\to\infty$, we have:
$$C=0$$
which contradicts the coefficients being nonzero!
A: If any pair of these are linearly independent, then $cy_1 \neq y_2$, for any $c$. So, for example, $ce^{ax}\neq e^{bx}$. 
Suppose the opposite, $ce^{ax}=e^{bx}$. Dividing by $e^{ax}$, $c=e^{(b-a)x}$. But $c$ is a constant, it can't be a function of $x$. So we have a contradiction (if $a\neq b$), so they are linearly indpendent, and it works the same way if you use $5$ instead of one of the exponential functions.
