Understanding and usage of Arrhenius equation
What activation energy would lead to a doubling in the reaction rate
  for a 10 degree increase in temperature, using the Arrhenius equation?


You’re right it is not generally true. Let’s set up an equation to solve the problem.

The Arrhenius equation is
$$k=A\operatorname{e}^{\frac{-E_\text{a}}{RT}}$$

In this case

$$2=\frac{A\operatorname{e}^{\frac{-E_\text{a}}{R\cdot(T + 10\ \mathrm{K})}}}{A\operatorname{e}^{\frac{-E_\text{a}}{RT}}}$$

or

$$\ln 2 = \frac{ - E_\text{a}}{R\left( T + 10\ \text{K} \right)} - \frac{ - E_\text{a}}{RT}$$

let’s pick a temperature around room temperature, 300 K is a nice even number, but you can re-solve the problem with any temperature you like.
$$0.69 = \frac{ - E_\text{a}}{R \cdot 310\ \text{K} } - \frac{ - E_\text{a}}{R \cdot 300\ \text{K} }$$
Using $R=8.31\ \mathrm{J/(K\ mol)}$ will produce an answer in J/mol.  Solving we find
$$E_\text{a} = 53\,568.49\ \mathrm{J/mol}$$
or,
$$53.6\ \mathrm{kJ/mol}~~(12.8\ \mathrm{kcal/mol})$$

So around room temperature a reaction that has an activation energy around 54 kJ/mol (or 13 kcal/mol) would show a doubling in rate for a 10 kelvin increase in temperature.