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

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

## See also

 2003 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 16 Followed byProblem 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

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