1951 AHSME Problems/Problem 35

Problem

If $a^{x}= c^{q}= b$ and $c^{y}= a^{z}= d$ , then

$\textbf{(A)}\ xy = qz\qquad\textbf{(B)}\ \frac{x}{y}=\frac{q}{z}\qquad\textbf{(C)}\ x+y = q+z\qquad\textbf{(D)}\ x-y = q-z$ $\textbf{(E)}\ x^{y}= q^{z}$

Solution

Try solving both equations for $a$ . Taking the $x$ -th root of both sides in the first equation and the $z$ -the root of both sides in the second gives $a=c^{\frac{q}x}$ and $a=c^{\frac{y}z}$ . So $\frac{q}x=\frac{y}z$ . Multiplying both sides by $xz$ , $qz=xy$ . $\boxed{\textbf{(A)}}$ .

1951 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 34	Followed by Problem 36
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 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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

Page

Toolbox

Search

1951 AHSME Problems/Problem 35

Problem

Solution

See Also