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\\

Revision as of 23:13, 4 April 2015

Problem

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

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

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 real solution to $y$, but there is imaginary.

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

$\textbf{(E)}\  \text{Only some of the above statements are correct}$. 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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All AHSME Problems and Solutions

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