1996 AHSME Problems/Problem 8

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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
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