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

Revision as of 20:11, 18 August 2011

If $3 = k\cdot 2^r$ and $15 = k\cdot 4^r$ , then $r =$

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

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 $\boxed{D}$ .

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

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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>
+<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==
+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>.
+The second equation gives <math>k = \frac{15}{4^r}</math>
+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>.
+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:
+<math>5 = 2^r</math>
+<math>r = \log_2 5</math>, which is option <math>\boxed{D}</math>.
 ==See also==
 {{AHSME box|year=1996|num-b=7|num-a=9}}