1996 AHSME Problems/Problem 8

Problem

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

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

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 $\boxed{D}$ .

Solution 2

Take corresponding logs and split up each equation to obtain:

$\log_2 3 = (\log_2 k) + (r)$

$\log_4 15 = (log_4 k) + (r)$

Then subtract the log from each side to isolate r:

$\log_2 (\frac{3}{k}) = r$

$\log_4 (\frac{15}{k}) = r$

Then set equalities and solve for k:

$\log_2 (\frac{3}{k}) = \log_4 (\frac{15}{k})$

$\frac{3}{k} = \sqrt{\frac{15}{k}}$

After solving we find that $k = \frac{3}{5}$ . Plugging into either of the equations and solving (easiest with equation 1) we find that $r = \log_2 5 \implies \boxed{D}$

1996 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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1996 AHSME Problems/Problem 8

Contents

Problem

Solution

Solution 2

See also