Difference between revisions of "1951 AHSME Problems/Problem 35"
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| + | ==Problem== | ||
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| + | If <math> a^{x}= c^{q}= b </math> and <math> c^{y}= a^{z}= d </math>, then | ||
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| + | <math> \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 </math> | ||
| + | <math> \textbf{(E)}\ x^{y}= q^{z} </math> | ||
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| + | ==Solution== | ||
| + | Try solving both equations for <math>a</math>. Taking the <math>x</math>-th root of both sides in the first equation and the <math>z</math>-the root of both sides in the second gives <math>a=c^{\frac{q}x}</math> and <math>a=c^{\frac{y}z}</math>. | ||
| + | So <math>\frac{q}x=\frac{y}z</math>. Multiplying both sides by <math>xz</math>, <math>qz=xy</math>. <math>\boxed{\textbf{(A)}}</math>. | ||
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== See Also == | == See Also == | ||
{{AHSME 50p box|year=1951|num-b=34|num-a=36}} | {{AHSME 50p box|year=1951|num-b=34|num-a=36}} | ||
Latest revision as of 14:28, 19 April 2014
Problem
If 
and 
, then
Solution
Try solving both equations for 
. Taking the 
-th root of both sides in the first equation and the 
-the root of both sides in the second gives 
and 
.
So 
. Multiplying both sides by 
, 
. 
.
See Also
| 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. 
