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 $\fbox{B}$

1967 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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1967 AHSME Problems/Problem 23

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