Difference between revisions of "1967 AHSME Problems/Problem 23"
(Created page with "== Problem == If <math>x</math> is real and positive and grows beyond all bounds, then <math>\log_3{(6x-5)}-\log_3{(2x+1)}</math> approaches: <math>\textbf{(A)}\ 0\qquad \textbf...") |
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== Solution == | == Solution == | ||
− | <math>\fbox{B}</math> | + | Since <math>\log_b x - \log_b y = \log_b \frac{x}{y}</math>, the expression is equal to <math>\log_3 \frac{6x - 5}{2x + 1}</math>. |
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+ | The expression <math>\frac{6x - 5}{2x + 1}</math> is equal to <math>3 - \frac{8}{2x + 1}</math>. As <math>x</math> gets large, the second term approaches <math>0</math>, and thus <math>\frac{6x - 5}{2x + 1}</math> approaches <math>3</math>. Thus, the expression approaches <math>\log_3 3</math>, which is <math>1</math>. | ||
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+ | Alternately, we divide the numerator and denominator of <math>\frac{6x - 5}{2x + 1}</math> by <math>x</math> to get <math>\frac{6 - \frac{5}{x}}{2 + \frac{1}{x}}</math>. As <math>x</math> grows large, both fractions approach <math>0</math>, leaving <math>\frac{6}{2} = 3</math>, and so the expression approaches <math>\log_3 3 = 1</math>. | ||
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+ | With either reasoning, the answer is <math>\fbox{B}</math> | ||
== See also == | == See also == |
Latest revision as of 23:48, 12 July 2019
Problem
If is real and positive and grows beyond all bounds, then approaches:
Solution
Since , the expression is equal to .
The expression is equal to . As gets large, the second term approaches , and thus approaches . Thus, the expression approaches , which is .
Alternately, we divide the numerator and denominator of by to get . As grows large, both fractions approach , leaving , and so the expression approaches .
With either reasoning, the answer is
See also
1967 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.