Difference between revisions of "1950 AHSME Problems/Problem 37"
m (→Solution) |
|||
Line 20: | Line 20: | ||
<math>\textbf{(D)}\ \text{If }0<x<a,y\text{ is always less than 0 and decreases without limit as }x\text{ approaches zero}</math>. Rewriting: <math>a^y=x</math> such that <math>x<a</math>. Well, a power of <math>a</math> can be less than <math>a</math> only if <math>y<1</math>. And we observe, <math>y</math> has no lower asymptote, because it is perfectly possible to have <math>y</math> be <math>-100000000</math>; in fact, the lower <math>y</math> gets, <math>x</math> approaches <math>0</math>. This is also correct. | <math>\textbf{(D)}\ \text{If }0<x<a,y\text{ is always less than 0 and decreases without limit as }x\text{ approaches zero}</math>. Rewriting: <math>a^y=x</math> such that <math>x<a</math>. Well, a power of <math>a</math> can be less than <math>a</math> only if <math>y<1</math>. And we observe, <math>y</math> has no lower asymptote, because it is perfectly possible to have <math>y</math> be <math>-100000000</math>; in fact, the lower <math>y</math> gets, <math>x</math> approaches <math>0</math>. This is also correct. | ||
− | <math>\textbf{(E)}\ \text{Only some of the above statements are correct}</math>. This is the last option, so | + | <math>\textbf{(E)}\ \text{Only some of the above statements are correct}</math>. This is the last option, so it follows that our answer is <math>\boxed{\textbf{(E)}\ \text{Only some of the above statements are correct}}</math> |
==See Also== | ==See Also== |
Revision as of 20:45, 6 July 2014
Problem
If $y \equal{} \log_{a}{x}$ (Error compiling LaTeX. Unknown error_msg), , which of the following statements is incorrect?
Solution
Let us first check
. Rewriting into exponential form gives . This is certainly correct.
. Rewriting gives . This is also certainly correct.
. Rewriting gives . Because , therefore positive, there is no real solution to , but there is imaginary.
. Rewriting: such that . Well, a power of can be less than only if . And we observe, has no lower asymptote, because it is perfectly possible to have be ; in fact, the lower gets, approaches . This is also correct.
. This is the last option, so it follows that our answer is
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 36 |
Followed by Problem 38 | |
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.