$8 \cdot 1$ satisfies the equation $2^{2 x}-8 \cdot 2^{x}+12=0$ for the value of $x$ as
(A) $\lg 3$.
(B) $\frac{1}{2} \lg 6$.
(C) $1+\lg \frac{3}{2}$.
(D) $1+\frac{\lg 3}{\lg 2}$.
(E) none of the above.
(20th American High School Mathematics Examination, 1969)
[Solution] The given equation is $\left(2^{x}\right)^{2}-8 \cdot 2^{x}+12=0$,
which gives $2^{x}=2$ or 6.
When $2^{x}=2$, $x=1$;
When $2^{x}=6$, we have $x \lg 2=\lg 2+\lg 3$, then $x=1+\frac{\lg 3}{\lg 2}$, hence the answer is $(E)$.