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 $xz$, $qz=xy$. $\boxed{\textbf{(A)}}$.

See Also

1951 AHSC (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS