# 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 $r = \log_2 5 \implies \boxed{D}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 