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

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It is not difficult to find <math>x = 10^{\frac{2}{5}}, y = 10^{\frac{1}{5}}</math>.
 
It is not difficult to find <math>x = 10^{\frac{2}{5}}, y = 10^{\frac{1}{5}}</math>.
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== Solution 2 ==
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<math>log(xy)+log(y^2)=1 \\ log(xy)+log(x)=1</math> subtracting, <math> \\ log(y^2)-log(x)=0 \\ log \left(\frac{y^2}{x}\right)=0 \\ y^2=x \\ \text{substitute and solve:} 5log(y)=1 \text{ and we need } 3log(y) \text{ which is } \frac{3}{5}</math>
  
 
== See also ==
 
== See also ==

Revision as of 20:18, 22 November 2019

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

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

See also

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

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