10. The function $y=a^{2 x}+2 a^{x}-1(a>0, a \neq 1)$ has a maximum value of 14 on the interval $[-1,1]$, the value of $a$ is $\qquad$ .
10. 3 or $\frac{1}{3} \cdot y=\left(a^{x}+1\right)^{2}-2, y_{\text {max }}=14$, then $a^{x}=3$, so $a^{-1}=3$ or $a^{1}=3$, thus $a=\frac{1}{3}$ or
3.