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

## 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 $$\frac{1000}{x^5} = 10$$ $$x = \sqrt{100} = 10^{\frac{2}{5}}$$ Solving for $y$, wee see it is equal to $10^{\frac{1}{5}}$.

Thus, $$\log_{10} xy = \frac{3}{5} \implies \boxed{\frac{3}{5}}$$

~YBSuburbanTea

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