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Difference between revisions of "1959 AHSME Problems/Problem 23"

(Created page with "== Problem 23== The set of solutions of the equation <math>\log_{10}\left( a^2-15a\right)=2</math> consists of <math>\textbf{(A)}\ \text{two integers } \qquad\textbf{(B)}\ \te...")
 
m (more polished, corrected inaccurate statement)
 
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<math>\textbf{(A)}\ \text{two integers } \qquad\textbf{(B)}\ \text{one integer and one fraction}\qquad \textbf{(C)}\ \text{two irrational numbers }\qquad\textbf{(D)}\ \text{two non-real numbers} \qquad\textbf{(E)}\ \text{no numbers, that is, the empty set}  </math>  
 
<math>\textbf{(A)}\ \text{two integers } \qquad\textbf{(B)}\ \text{one integer and one fraction}\qquad \textbf{(C)}\ \text{two irrational numbers }\qquad\textbf{(D)}\ \text{two non-real numbers} \qquad\textbf{(E)}\ \text{no numbers, that is, the empty set}  </math>  
 
== Solution ==
 
== Solution ==
−
Understand that <math>\log_{10}\left(x\right)=y</math> can be expressed as <math>x = y^2</math>.
+
Understand that <math>\log_{10}\left(x\right)=y</math> can be expressed as <math>x = 10^y</math>.
  
−
By applying that same logic to <math>\log_{10}\left( a^2-15a\right)=2</math>, we get <math>a^2-15a = 10^2</math>
+
By applying this fact to <math>\log_{10}\left( a^2-15a\right)=2</math>, we get <math>a^2-15a = 10^2</math>
 +
 
 +
We can use factoring methods to bring us to <math>(a-20)(a+5)=0</math>, which, as each binomial produces one integer solution, gives us a solution set of <math>\fbox{\textbf{(A) }two integers}</math>.
  
−
We can use factoring methods to bring us to <math>(a-20)(a+5)=0</math>, which, as each binomial produces one integer solution, gives us a set of <math>\textbf{(A)}</math> two integers in the solution set.
 
 
== See also ==
 
== See also ==
 
{{AHSME 50p box|year=1959|num-b=22|num-a=24}}
 
{{AHSME 50p box|year=1959|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:24, 21 July 2024

Problem 23

The set of solutions of the equation $\log_{10}\left( a^2-15a\right)=2$ consists of $\textbf{(A)}\ \text{two integers } \qquad\textbf{(B)}\ \text{one integer and one fraction}\qquad \textbf{(C)}\ \text{two irrational numbers }\qquad\textbf{(D)}\ \text{two non-real numbers} \qquad\textbf{(E)}\ \text{no numbers, that is, the empty set}$

Solution

Understand that $\log_{10}\left(x\right)=y$ can be expressed as $x = 10^y$.

By applying this fact to $\log_{10}\left( a^2-15a\right)=2$, we get $a^2-15a = 10^2$

We can use factoring methods to bring us to $(a-20)(a+5)=0$, which, as each binomial produces one integer solution, gives us a solution set of $\fbox{\textbf{(A) }two integers}$.

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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All AHSME Problems and Solutions

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