# 1967 AHSME Problems/Problem 23

## Problem

If $x$ is real and positive and grows beyond all bounds, then $\log_3{(6x-5)}-\log_3{(2x+1)}$ approaches: $\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ 4\qquad \textbf{(E)}\ \text{no finite number}$

## Solution

Since $\log_b x - \log_b y = \log_b \frac{x}{y}$, the expression is equal to $\log_3 \frac{6x - 5}{2x + 1}$.

The expression $\frac{6x - 5}{2x + 1}$ is equal to $3 - \frac{8}{2x + 1}$. As $x$ gets large, the second term approaches $0$, and thus $\frac{6x - 5}{2x + 1}$ approaches $3$. Thus, the expression approaches $\log_3 3$, which is $1$.

Alternately, we divide the numerator and denominator of $\frac{6x - 5}{2x + 1}$ by $x$ to get $\frac{6 - \frac{5}{x}}{2 + \frac{1}{x}}$. As $x$ grows large, both fractions approach $0$, leaving $\frac{6}{2} = 3$, and so the expression approaches $\log_3 3 = 1$.

With either reasoning, the answer is $\fbox{B}$

## See also

 1967 AHSME (Problems • Answer Key • Resources) Preceded byProblem 22 Followed byProblem 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 All AHSME Problems and Solutions

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