Difference between revisions of "1959 AHSME Problems/Problem 23"
Angrybird029 (talk | contribs) (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...") |
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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 | + | Understand that <math>\log_{10}\left(x\right)=y</math> can be expressed as <math>x = 10^y</math>. |
− | By applying | + | 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>. | ||
− | |||
== 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 12:24, 21 July 2024
Problem 23
The set of solutions of the equation consists of
Solution
Understand that can be expressed as .
By applying this fact to , we get
We can use factoring methods to bring us to , which, as each binomial produces one integer solution, gives us a solution set of .
See also
1959 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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