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

Solution 3

Since we have a system we can divide both equations: $\frac{15=k \cdot 4^r}{3=k \cdot 2^r}$. It is also important to know that $4^r$ is the same as $2^{2r}$. Simplifying the division of both equations you get $5=2^r$. Using the basic log formula $\log_m n = x \implies m^x=n$. Since $m^x=n$ plugging the numbers in makes $m=2$, $x=r$, $n=5$. This means $r$ is $\boxed{\textbf{(D)}\log_2 5}$

~AM24


See also

1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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