Why is the dual exponent what it is? The dual $p'$ of an exponent $p \in [0, \infty]$ is:
$$p'= \begin{cases} \frac{p}{p-1}, & 1 < p < \infty \\
\infty, & p=1 \\
1, & p= \infty. \end{cases}$$
Why is this the dual of an exponent?
 A: Suppose you want to write down an inequality
$$\|fg\|_1 \le C \|f\|_p \|g\|_q$$
for some $p, q$ and a universal constant $C$. The thing is that this inequality has to be homogeneous: dilations of $f$ shouldn't affect the inequality, nor should dilations of $g$ or dilations of the underlying measure. The fact that this is dilation invariant follows from the fact that $\|\lambda f\|_p = |\lambda| \|f\|_p$ for all $p$, which is why we take norms on both sides.
Dilating the measure gives us a relationship between $p$ and $q$. In particular, if we take $f = g = \chi_E$ to be the indicator of some measurable set, then the desired inequality reads
$$\mu(E) \le C \mu(E)^{1/p} \cdot \mu(E)^{1/q} = C \mu(E)^{1/p + 1/q}.$$
Now if $1/p + 1/q < 1$ and we have sets $E$ of arbitrarily large measure, there is no $C$ large enough to compensate for this and the inequality can't hold. Likewise, if $1/p + 1/q > 1$, we have a problem with sets of small measure.
So the happy medium is when $1/p + 1/q = 1$, which is exactly when $q = p'$. The moral of this story is that the full form Hölder's inequality can only hold with this choice of exponent, and that gives us the right dual space of $L^p$.

As a comment, homogeneity in various forms is a wonderful detector of false inequalities as well as a great predictor of true ones. It is one of my favorite techniques.
