Difference between revisions of "1996 AHSME Problems/Problem 8"

Latest revision as of 12:51, 16 August 2025

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 43: / Line 43: @@
 After solving we find that <math>k = \frac{3}{5}</math>. Plugging into either of the equations and solving (easiest with equation 1) we find that <math>r = \log_2 5 \implies \boxed{D}</math>
+==Solution 3==
+Since we have a system we can divide both equations: <math>\frac{15=k \cdot 4^r}{3=k \cdot 2^r}</math>. It is also important to know that <math>4^r</math> is the same as <math>2^{2r}</math>. Simplifying the division of both equations you get <math>5=2^r</math>. Using the basic log formula <math>\log_m n = x \implies m^x=n</math>. Since <math>m^x=n</math> plugging the numbers in makes <math>m=2</math>, <math>x=r</math>, <math>n=5</math>. This means <math>r</math> is <math>\boxed{\textbf{(D)}\log_2 5}</math>
+~[https://artofproblemsolving.com/wiki/index.php/User:Am24 AM24]

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