Quadratic or logarithmic?

Let $2014^x=a$ . We get $2014^{a^2+0.5}=2014^{2014}$ So $a^2-2014a+0.5=0$ If $x_1,x_2$ were the solutions to the original equation and $a_1,a_2$ for the second equation, $x_1\times x_2 = 2014^{a_1+a_1}=2014^{2014}$ The units digit of powers of 4 varies as $4,6,4,6...$ and 2014 is even so the answer is 6