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

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== Solution 2 ==
 
== Solution 2 ==
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<math>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}</math>
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<math>\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}</math>
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== Solution 3 ==
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Converting the two equation to exponential form, <math>\log_{10} xy^3 = 1 \implies 10 = xy^3</math> and <math>\log_{10} x^2y = 1 \implies 10 = x^2y</math>
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Solving for <math>y</math> in the second equation, <math>y = \frac{10}{x^2}</math>.
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Substituting this into the first equation, we see
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<cmath> \frac{1000}{x^5} = 10 </cmath>
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<cmath> x =  \sqrt[5]{100} = 10^{\frac{2}{5}} </cmath>
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Solving for <math>y</math>, wee see it is equal to  <math>10^{\frac{1}{5}}</math>.
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Thus, <cmath>\log_{10} xy = \frac{3}{5} \implies \boxed{\frac{3}{5}}</cmath>
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~YBSuburbanTea
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==Solution 4==
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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>.
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~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
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==Solution 5==
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Let <math>\log(x) = a, \log(y) = b</math>. The first equation can be written as <math>a + 3b = 1</math>, and the second as <math>2a + b = 1</math>. Solving this system of equations, we get that <math>a = \frac{2}{5}</math>, and <math>b = \frac{1}{5}</math>. Thus, the value of the expression we want to find is <math>a + b = \frac{1}{5} + \frac{2}{5} = \boxed{\textbf{(D)}~\frac{3}{5}}</math>
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~andliu766
  
 
== See also ==
 
== See also ==

Latest revision as of 13:05, 5 June 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

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 (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

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