Difference between revisions of "1965 AHSME Problems/Problem 31"

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== Solution ==
 
== Solution ==
<math>\fbox{C}</math>
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From the properties of [[logarithms]], we know that <math>(\log_ax)(\log_xb)=\log_ab</math>, and <math>\log_xb=\frac{1}{\log_bx}</math>. Thus, to satisfy the given equation, we are looking for values of <math>x</math> when <math>\log_bx=\frac{1}{\log_bx}</math>, or when <math>\log_bx=\pm 1</math>. This happens either when <math>x=b</math> or <math>x=\frac{1}{b}</math>. Thus we have <math>2</math> solutions, consistent with answer choice <math>\fbox{\textbf{(C)}}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 09:43, 19 July 2024

Problem

The number of real values of $x$ satisfying the equality $(\log_ax)(\log_bx) = \log_ab$, where $a > 0, b > 0, a \neq 1, b \neq 1$, is:

$\textbf{(A)}\ 0 \qquad  \textbf{(B) }\ 1 \qquad  \textbf{(C) }\ 2 \qquad  \textbf{(D) }\ \text{a finite integer greater than 2}\qquad \textbf{(E) }\ \text{not finite}$

Solution

From the properties of logarithms, we know that $(\log_ax)(\log_xb)=\log_ab$, and $\log_xb=\frac{1}{\log_bx}$. Thus, to satisfy the given equation, we are looking for values of $x$ when $\log_bx=\frac{1}{\log_bx}$, or when $\log_bx=\pm 1$. This happens either when $x=b$ or $x=\frac{1}{b}$. Thus we have $2$ solutions, consistent with answer choice $\fbox{\textbf{(C)}}$.

See Also

1965 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 30
Followed by
Problem 32
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
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