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

(Solution 1)
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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>
 
<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==
+
==Solution==
  
 
We want to find <math>r</math>, so our strategy is to eliminate <math>k</math>.
 
We want to find <math>r</math>, so our strategy is to eliminate <math>k</math>.
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<math>r = \log_2 5</math>, which is option <math>\boxed{D}</math>.
 
<math>r = \log_2 5</math>, which is option <math>\boxed{D}</math>.
 
  
 
==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 20:12, 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

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
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