1959 AHSME Problems/Problem 23

Revision as of 12:24, 21 July 2024 by Thepowerful456 (talk | contribs) (more polished, corrected inaccurate statement)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png