Difference between revisions of "2009 AIME II Problems/Problem 2"
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<cmath> a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}. </cmath> | <cmath> a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}. </cmath> | ||
− | == Solution == | + | == Solution 1 == |
− | |||
− | |||
First, we have: | First, we have: | ||
Line 53: | Line 51: | ||
and therefore the answer is <math>343+121+5 = \boxed{469}</math>. | and therefore the answer is <math>343+121+5 = \boxed{469}</math>. | ||
− | + | == Solution 2 == | |
+ | |||
+ | We know from the first three equations that <math>\log_a27 = \log_37</math>, <math>\log_b49 = \log_711</math>, and <math>\log_c\sqrt{11} = \log_{11}25</math>. Substituting, we find | ||
− | + | <cmath>a^{(\log_a27)(\log_37)} + b^{(\log_b49)(\log_711)} + c^{(\log_c\sqrt {11})(\log_{11}25)}.</cmath> | |
− | <math> | + | We know that <math>x^{\log_xy} =y</math>, so we find |
− | + | <cmath>27^{\log_37} + 49^{\log_711} + \sqrt {11}^{\log_{11}25}</cmath> | |
− | < | + | <cmath>(3^{\log_37})^3 + (7^{\log_711})^2 + ({11^{\log_{11}25}})^{1/2}.</cmath> |
− | <math> | + | The <math>3</math> and the <math>\log_37</math> cancel to make <math>7</math>, and we can do this for the other two terms. Thus, our answer is |
− | + | <cmath>7^3 + 11^2 + 25^{1/2}</cmath> | |
+ | <cmath>= 343 + 121 + 5</cmath> | ||
+ | <cmath>= \boxed {469}.</cmath> | ||
− | <math>7^3</math> | + | == Solution 3 == |
+ | First, let us take the log base 3 of the first expression. We get <math>\log_3{a^{\log_3{7}}} = 3</math>. Simplifying, we get <cmath>(\log_3{7})(\log_3{a}) = 3</cmath>. So, <cmath>\log_3 a = \frac{3}{\log_3{7}}</cmath>, and <cmath>a = 3^\frac{3}{log_3{7}}</cmath>. We can repeat the same process for the other equations, giving us <cmath>b = 7^\frac{2}{\log_7{11}}</cmath>, and <cmath>c = (\sqrt{11})^\frac{1}{\log_ | ||
+ | {11}{25}}</cmath>. Raising <math>a</math> to the power of <math>(\log_3{7})^2</math>, we get <cmath>3^{3\log_3{7}} = 3^{\log_3{343}} = 343</cmath>. Repeating a similar process for the other ones(you have to turn the square root to a fractional power for c), we get <math>343+121+5 = \boxed{469}</math> | ||
− | + | ~idk12345678 | |
− | |||
== See Also == | == See Also == | ||
{{AIME box|year=2009|n=II|num-b=1|num-a=3}} | {{AIME box|year=2009|n=II|num-b=1|num-a=3}} | ||
+ | [[Category: Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 15:20, 10 May 2024
Problem
Suppose that ,
, and
are positive real numbers such that
,
, and
. Find
Solution 1
First, we have:
Now, let , then we have:
This is all we need to evaluate the given formula. Note that in our case we have ,
, and
. We can now compute:
Similarly, we get
and
and therefore the answer is .
Solution 2
We know from the first three equations that ,
, and
. Substituting, we find
We know that , so we find
The and the
cancel to make
, and we can do this for the other two terms. Thus, our answer is
Solution 3
First, let us take the log base 3 of the first expression. We get . Simplifying, we get
. So,
, and
. We can repeat the same process for the other equations, giving us
, and
. Raising
to the power of
, we get
. Repeating a similar process for the other ones(you have to turn the square root to a fractional power for c), we get
~idk12345678
See Also
2009 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.