Difference between revisions of "1950 AHSME Problems/Problem 37"

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==Problem==
 
==Problem==
  
If <math> y \equal{} \log_{a}{x}</math>, <math> a > 1</math>, which of the following statements is incorrect?
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If <math> y = \log_{a}{x}</math>, <math> a > 1</math>, which of the following statements is incorrect?
  
 
<math>\textbf{(A)}\ \text{If }x=1,y=0 \qquad\\
 
<math>\textbf{(A)}\ \text{If }x=1,y=0 \qquad\\
 
\textbf{(B)}\ \text{If }x=a,y=1 \qquad\\
 
\textbf{(B)}\ \text{If }x=a,y=1 \qquad\\
 
\textbf{(C)}\ \text{If }x=-1,y\text{ is imaginary (complex)} \qquad\\
 
\textbf{(C)}\ \text{If }x=-1,y\text{ is imaginary (complex)} \qquad\\
\textbf{(D)}\ \text{If }0<x<a,y\text{ is always less than 0 and decreases without limit as }x\text{ approaches zero} \qquad\\
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\textbf{(D)}\ \text{If }0<x<1,y\text{ is always less than 0 and decreases without limit as }x\text{ approaches zero} \qquad\\
 
\textbf{(E)}\  \text{Only some of the above statements are correct}</math>
 
\textbf{(E)}\  \text{Only some of the above statements are correct}</math>
  
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<math>\textbf{(C)}\ \text{If }x=-1,y\text{ is imaginary (complex)}</math>. Rewriting gives <math>a^{\text{complex number}}=-1</math>. Because <math>a>1</math>, therefore positive, there is no real solution to <math>y</math>, but there is imaginary.
 
<math>\textbf{(C)}\ \text{If }x=-1,y\text{ is imaginary (complex)}</math>. Rewriting gives <math>a^{\text{complex number}}=-1</math>. Because <math>a>1</math>, therefore positive, there is no real solution to <math>y</math>, but there is imaginary.
  
<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.
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<math>\textbf{(D)}\ \text{If }0<x<1,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 ti follows that our answer is <math>\boxed{\textbf{(E)}\  \text{Only some of the above statements are correct}}</math>
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<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==

Latest revision as of 17:09, 15 March 2017

Problem

If $y = \log_{a}{x}$, $a > 1$, which of the following statements is incorrect?

$\textbf{(A)}\ \text{If }x=1,y=0 \qquad\\ \textbf{(B)}\ \text{If }x=a,y=1 \qquad\\ \textbf{(C)}\ \text{If }x=-1,y\text{ is imaginary (complex)} \qquad\\ \textbf{(D)}\ \text{If }0<x<1,y\text{ is always less than 0 and decreases without limit as }x\text{ approaches zero} \qquad\\ \textbf{(E)}\  \text{Only some of the above statements are correct}$

Solution

Let us first check

$\textbf{(A)}\ \text{If }x=1,y=0$. Rewriting into exponential form gives $a^0=1$. This is certainly correct.

$\textbf{(B)}\ \text{If }x=a,y=1$. Rewriting gives $a^1=a$. This is also certainly correct.

$\textbf{(C)}\ \text{If }x=-1,y\text{ is imaginary (complex)}$. Rewriting gives $a^{\text{complex number}}=-1$. Because $a>1$, therefore positive, there is no real solution to $y$, but there is imaginary.

$\textbf{(D)}\ \text{If }0<x<1,y\text{ is always less than 0 and decreases without limit as }x\text{ approaches zero}$. Rewriting: $a^y=x$ such that $x<a$. Well, a power of $a$ can be less than $a$ only if $y<1$. And we observe, $y$ has no lower asymptote, because it is perfectly possible to have $y$ be $-100000000$; in fact, the lower $y$ gets, $x$ approaches $0$. This is also correct.

$\textbf{(E)}\  \text{Only some of the above statements are correct}$. This is the last option, so it follows that our answer is $\boxed{\textbf{(E)}\  \text{Only some of the above statements are correct}}$

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 36
Followed by
Problem 38
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