1996 AHSME Problems/Problem 8

Problem

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

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

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
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