10.86 The value of $x$ that satisfies the equation $x^{x^{x}}=2$ is
(A) infinity.
(B) 2.
(C) $4 \sqrt{2}$.
(D) $\sqrt{2}$.
(E) None of the above answers is correct.
(1st American High School Mathematics Examination, 1950)
[Solution] From the given, we have $\lg x^{x^{x}}=\lg 2$, which means $x^{x^{2}} \lg x=\lg 2$. But $x^{x^{x^{x}}}=2$, so $2 \lg x=\lg 2$, which implies $x^{2}=2$. Thus, $x= \pm \sqrt{2}$ (discard the negative value). Therefore, the answer is $(D)$.