# 1969 AHSME Problems/Problem 29

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

If $x=t^{1/(t-1)}$ and $y=t^{t/(t-1)},t>0,t \ne 1$, a relation between $x$ and $y$ is:

$\text{(A) } y^x=x^{1/y}\quad \text{(B) } y^{1/x}=x^{y}\quad \text{(C) } y^x=x^y\quad \text{(D) } x^x=y^y\quad \text{(E) none of these}$

## Solution

Plug in $t = 2$ and test each expression (if the relation works, then both sides are equal). After testing, options A, B, and D are out, but for option C, both sides are equal. To check that C is a valid option, substitute the values of $x$ and $y$ and use exponent properties. $$x^y = (t^{\frac{1}{t-1}})^{t^{\frac{t}{t-1}}} = t^{\frac{1}{t-1} \cdot t^{\frac{t}{t-1}}}$$ $$y^x = (t^{\frac{t}{t-1}})^{t^{\frac{1}{t-1}}} = t^{\frac{t}{t-1} \cdot t^{\frac{1}{t-1}}}$$ In the second equation, the exponent can be rewritten as $\tfrac{1}{t-1} \cdot t \cdot t^{\frac{1}{t-1}}$. Since $\tfrac{t-1}{t-1} = 1$, the exponent can be simplified as $\tfrac{1}{t-1} \cdot t^{\frac{t-1}{t-1}} \cdot t^{\frac{1}{t-1}} = \tfrac{1}{t-1} \cdot t^{\frac{t}{t-1}}$. Since the base and exponent of both $x^y$ and $y^x$ are the same, we can confirm that $\boxed{\textbf{(C) } y^x = x^y}$.

## See Also

 1969 AHSC (Problems • Answer Key • Resources) Preceded byProblem 28 Followed byProblem 30 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Invalid username
Login to AoPS