1967 AHSME Problems/Problem 23

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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 (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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