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 \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> | <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> | ||
== See also == | == See also == |
Revision as of 11:53, 14 January 2022
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,
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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