Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2003 AMC 12B Problems/Problem 17"

(→‎Solution 2)
(→‎Solution 2)
Line 22: Line 22:
 
== Solution 2 ==
 
== Solution 2 ==
 
<math>\log(xy)+\log(y^2)=1 \\ \log(xy)+\log(x)=1 \text{ subtracting, } \\ \log(y^2)-\log(x)=0 \\ \log \left(\frac{y^2}{x}\right)=0 \\ \frac{y^2}{x}=10^0 \\ y^2=x \\ \text{substitute and solve: } \log(y^5)=5\log(y)=1 \\ \text{ and we need } 3\log(y) \text{ which is } \frac{3}{5}</math>
 
<math>\log(xy)+\log(y^2)=1 \\ \log(xy)+\log(x)=1 \text{ subtracting, } \\ \log(y^2)-\log(x)=0 \\ \log \left(\frac{y^2}{x}\right)=0 \\ \frac{y^2}{x}=10^0 \\ y^2=x \\ \text{substitute and solve: } \log(y^5)=5\log(y)=1 \\ \text{ and we need } 3\log(y) \text{ which is } \frac{3}{5}</math>
 +
 +
== Solution 3 ==
 +
 +
Converting the two equation to exponential form, <math>\log_{10} xy^3 = 1 \implies 10 = xy^3</math> and <math>\log_{10} x^2y = 1 \implies 10 = x^2y</math>
 +
\\
 +
Solving for <math>y</math> in the second equation, <math>y = \frac{10}{x^2}</math>.
 +
\\
 +
Substituting this into the first equation, we see
 +
<cmath> \frac{1000}{x^5} = 10 </cmath>
 +
<cmath> x =  \sqrt[5]{100} = 10^{\frac{2}{5}} </cmath>
 +
Solving for <math>y</math>, wee see it is equal to  <math>10^{\frac{1}{5}}</math>.
 +
\\
 +
Thus, <cmath>\log_{10} xy = \frac{3}{5} \implies \boxed{\frac{3}{5}}</cmath>
  
 
== See also ==
 
== See also ==

Revision as of 10:53, 14 January 2022

Problem

If $\log (xy^3) = 1$ and $\log (x^2y) = 1$, what is $\log (xy)$?

$\mathrm{(A)}\ -\frac 12 \qquad\mathrm{(B)}\ 0 \qquad\mathrm{(C)}\ \frac 12 \qquad\mathrm{(D)}\ \frac 35 \qquad\mathrm{(E)}\ 1$

Solution

Since \begin{align*} &\log(xy) +2\log y = 1 \\ \log(xy) + \log x = 1 \quad \Longrightarrow \quad &2\log(xy) + 2\log x = 2 \end{align*} Summing gives \[3\log(xy) + 2\log y + 2\log x = 3 \Longrightarrow 5\log(xy) = 3\]

Hence $\log (xy) = \frac 35 \Rightarrow \mathrm{(D)}$.

It is not difficult to find $x = 10^{\frac{2}{5}}, y = 10^{\frac{1}{5}}$.

Solution 2

$\log(xy)+\log(y^2)=1 \\ \log(xy)+\log(x)=1 \text{ subtracting, } \\ \log(y^2)-\log(x)=0 \\ \log \left(\frac{y^2}{x}\right)=0 \\ \frac{y^2}{x}=10^0 \\ y^2=x \\ \text{substitute and solve: } \log(y^5)=5\log(y)=1 \\ \text{ and we need } 3\log(y) \text{ which is } \frac{3}{5}$

Solution 3

Converting the two equation to exponential form, $\log_{10} xy^3 = 1 \implies 10 = xy^3$ and $\log_{10} x^2y = 1 \implies 10 = x^2y$ \\ Solving for $y$ in the second equation, $y = \frac{10}{x^2}$. \\ Substituting this into the first equation, we see \[\frac{1000}{x^5} = 10\] \[x = \sqrt[5]{100} = 10^{\frac{2}{5}}\] Solving for $y$, wee see it is equal to $10^{\frac{1}{5}}$. \\ Thus, \[\log_{10} xy = \frac{3}{5} \implies \boxed{\frac{3}{5}}\]

See also

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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
All AMC 12 Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12B_Problems/Problem_17&oldid=169878"