Difference between revisions of "1996 AHSME Problems/Problem 8"
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+ | ==Problem== | ||
+ | |||
+ | If <math>3 = k\cdot 2^r</math> and <math>15 = k\cdot 4^r</math>, then <math>r = </math> | ||
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+ | <math> \text{(A)}\ -\log_{2}5\qquad\text{(B)}\ \log_{5}2\qquad\text{(C)}\ \log_{10}5\qquad\text{(D)}\ \log_{2}5\qquad\text{(E)}\ \frac{5}{2} </math> | ||
+ | |||
+ | ==Solution 1== | ||
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+ | We want to find <math>r</math>, so our strategy is to eliminate <math>k</math>. | ||
+ | |||
+ | The first equation gives <math>k = \frac{3}{2^r}</math>. | ||
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+ | The second equation gives <math>k = \frac{15}{4^r}</math> | ||
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+ | Setting those two equal gives <math>\frac{3}{2^r} = \frac{15}{4^r}</math> | ||
+ | |||
+ | Cross-multiplying and dividing by <math>3</math> gives <math>5\cdot 2^r = 4^r</math>. | ||
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+ | We know that <math>4^r = 2^r \cdot 2^r</math>, so we can divide out <math>2^r</math> from both sides (which is legal since <math>2^r \neq 0</math>), and we get: | ||
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+ | <math>5 = 2^r</math> | ||
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+ | <math>r = \log_2 5</math>, which is option <math>\boxed{D}</math>. | ||
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==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=7|num-a=9}} | {{AHSME box|year=1996|num-b=7|num-a=9}} |
Revision as of 20:11, 18 August 2011
Problem
If and , then
Solution 1
We want to find , so our strategy is to eliminate .
The first equation gives .
The second equation gives
Setting those two equal gives
Cross-multiplying and dividing by gives .
We know that , so we can divide out from both sides (which is legal since ), and we get:
, which is option .
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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 |