My (apparently wrong) reasoning that tells me the set is actually linearly dependent goes something like this:
Take the equation $a_1e^x + a_2e^{2x} = 0$. Assume that $a_1$ doesn't equal 0. Then we have the equation:
$$\frac{-a_2}{a_1}e^{2x} = e^x \tag{$*$} $$
For all $x$ we can find $a_1, a_2,$ satisfying this equation (we set $a_2 = -1, a_1 = e^x$ for whatever particular $x$ value is being evaluated).
Is the flaw in my reasoning the fact $a_1$ is dependent on different $x$'s for the equation to work for all $x$'s? If so, to prove that the set was linearly independent, would I just have to remark that there are no constants $a_1, a_2$ that make $(*)$ work for all $x$'s? , and hence contra the assumption, $a_1 = a_2 = 0$?
Best Answer
Linear dependence doesn't make sense without specifying what the scalars are. If you're allowed to use coefficients that are, say, continuous functions, then $\{ e^x,e^{2x} \}$ is, in fact, linearly dependent, by your very argument: you have a nonzero linear combination
$$ e^x \cdot e^x - 1 \cdot e^{2x} = 0 $$
giving zero.
However, if scalars are restricted to being just real numbers, then the linear combination above doesn't work to show dependence, because $e^x$ is not a scalar.
There is a simplification and an abuse of notation going on here that may be confusing you. Strictly speaking, $e^x$ is a real number (that varies depending on $x$), but the question intends to ask about a function.
Let me write $f$, $g$, $h$, and $k$ for the four functions defined by
$$ f(x) = e^{2x} \qquad g(x) = e^x \qquad h(x) = 1 \qquad k(x) = 0$$
The question is asking to show that $\{ f, g \}$ is a linearly independent set. I assume we are in the case that scalars are real numbers. So, the question is whether or not there not exist scalars $a,b$, such that
$$af + bg = k $$
(note that $k$ is the zero vector) Now, an equation of functions holds if and only if it holds for all values — so the problem is equivalent to asking of there are scalars $a$ and a $b$ such that, for every $x$, we have
$$ a f(x) + b g(x) = k(x) $$
or equivalently,
$$ a e^{2x} + b e^x = 0 $$
Since you can't find particular scalars $a$ and $b$ that make this equation true for all $x$, the functions are independent.
However, if we take the question literally without recognizing the intended abuse of notation, it is correct to say that the (variable) set of real numbers $\{ e^{2x}, e^x \}$ is linearly dependent (for all values of $x$), by the argument you gave. The thing we need to show has the quantification the other way around: the problem is, for each x, to find an $a$ and a $b$.