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 \[2 \cdot \textbf{(2)} + \textbf{(1)} = 5u + 5v = 3 \implies u + v = \frac{3}{5} \implies c = \boxed{\textbf{(D)}~\frac{3}{5}}\].

~ cxsmi

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

See also

2003 AMC 12B (ProblemsAnswer KeyResources)
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