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

Revision as of 12:49, 4 April 2024

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

@@ Line 39: / Line 39: @@
 ==Solution 4==
-We rewrite the logarithms in the problem. <cmath>\log(x) + 3\log(y) = 1</cmath> <cmath>2\log(x) + \log(y) = 1</cmath> <cmath>\log(x) + \log(y) = c</cmath> where <math>c</math> is the desired quantity. Set <math>u = \log(x)</math> and <math>v = \log(y)</math>. Then we have that <cmath>u + 3y = 1 \textbf{(1)}</cmath> <cmath>2u + y = 1 \textbf{(2)}</cmath> <cmath>u + v = c</cmath>. Notice that <cmath>2 \cdot \textbf{(2)} + \textbf{(1)} = 5u + 5v = 3 \implies u + v = \frac{3}{5} \implies c = \boxed{\textbf{(D)} \frac{3}{5}}</cmath>.
+We rewrite the logarithms in the problem. <cmath>\log(x) + 3\log(y) = 1</cmath> <cmath>2\log(x) + \log(y) = 1</cmath> <cmath>\log(x) + \log(y) = c</cmath> where <math>c</math> is the desired quantity. Set <math>u = \log(x)</math> and <math>v = \log(y)</math>. Then we have that <cmath>u + 3y = 1 \textbf{(1)}</cmath> <cmath>2u + y = 1 \textbf{(2)}</cmath> <cmath>u + v = c</cmath>. Notice that <cmath>2 \cdot \textbf{(2)} + \textbf{(1)} = 5u + 5v = 3 \implies u + v = \frac{3}{5} \implies c = \boxed{\textbf{(D)}~\frac{3}{5}}</cmath>.
-~ cxsmi
+~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
 == See also ==

2003 AMC 12B (Problems • Answer Key • Resources)
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

Page

Toolbox

Search