Difference between revisions of "1996 AHSME Problems/Problem 8"

(Created page with "==See also== {{AHSME box|year=1996|num-b=7|num-a=9}}")
 
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==Problem==
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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>
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==Solution 1==
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We want to find <math>r</math>, so our strategy is to eliminate <math>k</math>.
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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>
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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 $3 = k\cdot 2^r$ and $15 = k\cdot 4^r$, then $r =$

$\text{(A)}\ -\log_{2}5\qquad\text{(B)}\ \log_{5}2\qquad\text{(C)}\ \log_{10}5\qquad\text{(D)}\ \log_{2}5\qquad\text{(E)}\ \frac{5}{2}$

Solution 1

We want to find $r$, so our strategy is to eliminate $k$.

The first equation gives $k = \frac{3}{2^r}$.

The second equation gives $k = \frac{15}{4^r}$

Setting those two equal gives $\frac{3}{2^r} = \frac{15}{4^r}$

Cross-multiplying and dividing by $3$ gives $5\cdot 2^r = 4^r$.

We know that $4^r = 2^r \cdot 2^r$, so we can divide out $2^r$ from both sides (which is legal since $2^r \neq 0$), and we get:

$5 = 2^r$

$r = \log_2 5$, which is option $\boxed{D}$.


See also

1996 AHSME (ProblemsAnswer KeyResources)
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
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