6.21 Given $x=t^{\frac{1}{t-1}}$ and $y=t^{\frac{t}{t-1}}(t>0, t \neq 1)$, the relationship between $x$ and $y$ is
(A) $y^{x}=x^{\frac{1}{y}}$.
(B) $y^{\frac{1}{x}}=x^{y}$.
(C) $y^{x}=x^{y}$.
(D) $x^{x}=y^{y}$.
(E) None of the above.
(20th American High School Mathematics Examination, 1969)
［Solution 1］According to the given and noting that
$$
\frac{y}{x}=\frac{t^{\frac{t}{t-1}}}{\frac{1}{t-1}}=t^{\frac{t}{t-1}-\frac{1}{t-1}}=t,
$$

On the other hand,
$$
y=t^{\frac{t}{t-1}}=\left(t^{\frac{1}{t-1}}\right)^{t}=x^{t},
$$

Therefore, $y=x^{t}=x^{\frac{y}{x}}$, which means $y^{x}=x^{y}$. Hence, the answer is (C).
[Solution 2] From the above, we know $y=t x$, then $\frac{1}{t-1} y=\frac{t}{t-1} x$, i.e., $\left(\log _{x} x\right) y=\left(\log _{\iota} y\right) x$, or $\log _{x} x^{y}=\log _{2} y^{x}$,
$$
\therefore x^{y}=y^{x} \quad(\because t>0, \quad t \neq 1) \text {. }
$$

Hence, the answer is (C).