3. The solution to the equation $\arctan 2^{x}-\arctan 2^{-x}=\frac{\pi}{6}$
is . $\qquad$
3. $\log _{2} \sqrt{3}$.

Taking the tangent of both sides of the known equation, we get
$$
\begin{array}{l}
\frac{2^{x}-2^{-x}}{1+2^{x} \times 2^{-x}}=\frac{\sqrt{3}}{3} \Rightarrow 2^{x}-2^{-x}=\frac{2 \sqrt{3}}{3} \\
\Rightarrow\left(2^{x}\right)^{2}-\frac{2 \sqrt{3}}{3} \times 2^{x}-1=0 .
\end{array}
$$

Since $2^{x}>0$, we have
$$
2^{x}=\sqrt{3} \Rightarrow x=\log _{2} \sqrt{3} .
$$