1975 AHSME Problems/Problem 19

Revision as of 13:21, 26 December 2020 by Jiang147369 (talk | contribs) (Created page with "== Problem 19 == Which positive numbers <math>x</math> satisfy the equation <math>(\log_3x)(\log_x5)=\log_35</math>? <math>\textbf{(A)}\ 3 \text{ and } 5 \text{ only} \qqua...")
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Problem 19

Which positive numbers $x$ satisfy the equation $(\log_3x)(\log_x5)=\log_35$?

$\textbf{(A)}\ 3 \text{ and } 5 \text{ only} \qquad  \textbf{(B)}\ 3, 5, \text{ and } 15 \text{ only} \qquad \\ \textbf{(C)}\ \text{only numbers of the form } 5^n \cdot 3^m, \text{ where } n \text{ and } m \text{ are positive integers} \qquad \\ \textbf{(D)}\ \text{all positive } x \neq 1 \qquad  \textbf{(E)}\ \text{none of these}$

Solution

By the change-of-base formula, we can simplify the left side of the equation: $(\log_3x)(\log_x5) = (\frac{\log_x}{\log_3})(\frac{\log_5}{\log_x}) = \frac{\log_5}{\log_3}$.

We see that this in fact simplifies to $\log_35$, which will always equal the right side of the equation, since they are the same exact expressions.

But we have to be careful because $x \neq 1$. Plugging in $x=1$, the left side would equal $(\log_31)(\log_x5) = 0 \cdot \log_x5 = 0$, and $\log_35$ definitely does not equal $0$.

Besides $1$, $x$ can take on any positive value, and the equation would work. Therefore, the answer is $\boxed{\textbf{(D)}\ \text{all positive } x \neq 1}$ ~jiang147369

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