Difference between revisions of "2003 AMC 12B Problems/Problem 17"
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== Solution 2 == | == Solution 2 == | ||
− | <math>log(xy)+log(y^2)=1 \\ log(xy)+log(x)=1 | + | <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> |
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | Substituting this into the first equation, we see | ||
+ | <cmath> \frac{1000}{x^5} = 10 </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>. | ||
+ | |||
+ | Thus, <cmath>\log_{10} xy = \frac{3}{5} \implies \boxed{\frac{3}{5}}</cmath> | ||
+ | |||
+ | ~YBSuburbanTea | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | ~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] | ||
+ | |||
+ | ==Solution 5== | ||
+ | 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> | ||
+ | |||
+ | ~andliu766 | ||
== See also == | == See also == |
Latest revision as of 14:05, 5 June 2024
Problem
If and
, what is
?
Solution
Since
Summing gives
Hence .
It is not difficult to find .
Solution 2
Solution 3
Converting the two equation to exponential form, and
Solving for in the second equation,
.
Substituting this into the first equation, we see
Solving for
, wee see it is equal to
.
Thus,
~YBSuburbanTea
Solution 4
We rewrite the logarithms in the problem.
where
is the desired quantity. Set
and
. Then we have that
. Notice that
.
~ cxsmi
Solution 5
Let . The first equation can be written as
, and the second as
. Solving this system of equations, we get that
, and
. Thus, the value of the expression we want to find is
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.