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

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==Probelm==
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==Problem==
  
 
If <math> y \equal{} \log_{a}{x}</math>, <math> a > 1</math>, which of the following statements is incorrect?
 
If <math> y \equal{} \log_{a}{x}</math>, <math> a > 1</math>, which of the following statements is incorrect?
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\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<z,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<a,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>
  
 
==Solution==
 
==Solution==
{{solution}}
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Let us first check
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<math>\textbf{(A)}\ \text{If }x=1,y=0</math>. Rewriting into exponential form gives <math>a^0=1</math>. This is certainly correct.
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<math>\textbf{(B)}\ \text{If }x=a,y=1</math>. Rewriting gives <math>a^1=a</math>. This is also certainly correct.
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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 solution to <math>y</math>. Is it possible to have complex powers? Let's use process of elimination to find out the right answer.
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<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{(E)}\  \text{Only some of the above statements are correct}</math>. Now, we need to use logic to see if this is true or not. Assume that this is true; but that would mean that <math>\textbf{(C)}</math> and <math>\textbf{(E)}</math> would both be answers. This is definitely not possible, so it follows that <math>\textbf{(E)}</math> is incorrect, therefore the answer is <math>\boxed{\textbf{(E)}\  \text{Only some of the above statements are correct}}</math>
  
 
==See Also==
 
==See Also==

Revision as of 19:15, 10 April 2013

Problem

If $y \equal{} \log_{a}{x}$ (Error compiling LaTeX. Unknown error_msg), $a > 1$, which of the following statements is incorrect?

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

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 solution to $y$. Is it possible to have complex powers? Let's use process of elimination to find out the right answer.

$\textbf{(D)}\ \text{If }0<x<a,y\text{ is always less than 0 and decreases without limit as }x\text{ approaches zero}$. 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.

$\textbf{(E)}\  \text{Only some of the above statements are correct}$. Now, we need to use logic to see if this is true or not. Assume that this is true; but that would mean that $\textbf{(C)}$ and $\textbf{(E)}$ would both be answers. This is definitely not possible, so it follows that $\textbf{(E)}$ is incorrect, therefore the 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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All AHSME Problems and Solutions