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

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

~YBSuburbanTea

## Solution 4

We rewrite the logarithms in the problem. $$\log(x) + 3\log(y) = 1$$ $$2\log(x) + \log(y) = 1$$ $$\log(x) + \log(y) = c$$ where $c$ is the desired quantity. Set $u = \log(x)$ and $v = \log(y)$. Then we have that $$u + 3y = 1 \textbf{(1)}$$ $$2u + y = 1 \textbf{(2)}$$ $$u + v = c$$. Notice that $$2 \cdot \textbf{(2)} + \textbf{(1)} = 5u + 5v = 3 \implies u + v = \frac{3}{5} \implies c = \boxed{\textbf{(D)}~\frac{3}{5}}$$.

## Solution 5

Let $\log(x) = a, \log(y) = b$. The first equation can be written as $a + 3b = 1$, and the second as $2a + b = 1$. Solving this system of equations, we get that $a = \frac{2}{5}$, and $b = \frac{1}{5}$. Thus, the value of the expression we want to find is $a + b = \frac{1}{5} + \frac{2}{5} = \boxed{\textbf{(D)}~\frac{3}{5}}$

~andliu766