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2003 AMC 12B Problems/Problem 17: Difference between revisions

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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>
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>
\\
 
Solving for <math>y</math> in the second equation, <math>y = \frac{10}{x^2}</math>.
Solving for <math>y</math> in the second equation, <math>y = \frac{10}{x^2}</math>.
\\
 
Substituting this into the first equation, we see
Substituting this into the first equation, we see
<cmath> \frac{1000}{x^5} = 10 </cmath>
<cmath> \frac{1000}{x^5} = 10 </cmath>
<cmath> x =  \sqrt[5]{100} = 10^{\frac{2}{5}} </cmath>
<cmath> x =  \sqrt[5]{100} = 10^{\frac{2}{5}} </cmath>
Solving for <math>y</math>, wee see it is equal to  <math>10^{\frac{1}{5}}</math>.
Solving for <math>y</math>, wee see it is equal to  <math>10^{\frac{1}{5}}</math>.
\\
 
Thus, <cmath>\log_{10} xy = \frac{3}{5} \implies \boxed{\frac{3}{5}}</cmath>
Thus, <cmath>\log_{10} xy = \frac{3}{5} \implies \boxed{\frac{3}{5}}</cmath>
~YBSuburbanTea


== See also ==
== See also ==

Revision as of 10:54, 14 January 2022

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

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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