Difference between revisions of "2009 AIME II Problems/Problem 2"
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'''Solution 2''' | '''Solution 2''' | ||
| − | We know from the first three equations that <math>log_a27</math> = <math>log_37</math>, <math>log_b49</math> = <math>log_711</math>, and <math>log_c\sqrt{11}</math> = <math>log_{11}25</math>. Substituting, we get | + | We know from the first three equations that <math>\log_a27</math> = <math>\log_37</math>, <math>\log_b49</math> = <math>\log_711</math>, and <math>\log_c\sqrt{11}</math> = <math>\log_{11}25</math>. Substituting, we get |
| − | <math>a^{(log_a27)(log_37)}</math> + <math>b^{(log_b49)(log_711)</math> + <math>c^{(log_c\sqrt {11})(log_{11}25)}</math> | + | <math>a^{(\log_a27)(\log_37)}</math> + <math>b^{(\log_b49)(\log_711)</math> + <math>c^{(\log_c\sqrt {11})(\log_{11}25)}</math> |
| − | We know that <math>x^{log_xy}</math> = <math>y</math>, so we get | + | We know that <math>x^{\log_xy}</math> = <math>y</math>, so we get |
| − | <math>27^{log_37}</math> + <math>49^{log_711}</math> + <math>\sqrt {11}^{log_{11}25}</math> | + | <math>27^{\log_37}</math> + <math>49^{\log_711}</math> + <math>\sqrt {11}^{\log_{11}25}</math> |
| − | <math>(3^{log_37})^3</math> + <math>(7^{log_711})^2</math> + <math>({11^{log_{11}25})^{1/2}</math> | + | <math>(3^{\log_37})^3</math> + <math>(7^{\log_711})^2</math> + <math>({11^{\log_{11}25})^{1/2}</math> |
| − | The <math>3</math> and the <math>log_37</math> cancel out to make <math>7</math>, and we can do this for the other two terms. We obtain | + | The <math>3</math> and the <math>\log_37</math> cancel out to make <math>7</math>, and we can do this for the other two terms. We obtain |
<math>7^3</math> + <math>11^2</math> + <math>25^{1/2}</math> | <math>7^3</math> + <math>11^2</math> + <math>25^{1/2}</math> | ||
Revision as of 22:19, 9 June 2010
Problem
Suppose that 
, 
, and 
are positive real numbers such that 
, 
, and 
. Find
![\[a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}.\]](http://latex.artofproblemsolving.com/6/8/0/680c5fe965e912d7a3cbf4cf823fe104d86d3124.png)
Solution
Solution 1
First, we have:
![\[x^{(\log_y z)^2} = x^{\left( (\log_y z)^2 \right) } = x^{(\log_y z) \cdot (\log_y z) } = \left( x^{\log_y z} \right)^{\log_y z}\]](http://latex.artofproblemsolving.com/f/7/6/f76c9c999e1420dd3c53d8ae7234693c3ee3ab57.png)
Now, let 
, then we have:
![\[x^{\log_y z} = \left( y^w \right)^{\log_y z} = y^{w\log_y z} = y^{\log_y (z^w)} = z^w\]](http://latex.artofproblemsolving.com/f/9/b/f9b1389de58394163d7255c83166d35a6cafd033.png)
This is all we need to evaluate the given formula. Note that in our case we have 
, 
, and 
. We can now compute:
![\[a^{(\log_3 7)^2} = \left( a^{\log_3 7} \right)^{\log_3 7} = 27^{\log_3 7} = (3^3)^{\log_3 7} = 7^3 = 343\]](http://latex.artofproblemsolving.com/5/f/c/5fcca61e8df21a948b018cd8a897d88526c284aa.png)
Similarly, we get
![\[b^{(\log_7 11)^2} = (7^2)^{\log_7 11} = 11^2 = 121\]](http://latex.artofproblemsolving.com/a/e/2/ae280c7340770d1e8ddb9a87ab0807a078e19641.png)
and
![\[c^{(\log_{11} 25)^2} = (11^{1/2})^{\log_{11} 25} = 25^{1/2} = 5\]](http://latex.artofproblemsolving.com/a/a/7/aa762479266a6f6fef5325e9b7d2a7ea1c36f623.png)
and therefore the answer is 
.
Solution 2
We know from the first three equations that 
= 
, 
= 
, and 
= 
. Substituting, we get
+ $b^{(\log_b49)(\log_711)$ (Error compiling LaTeX. Unknown error_msg) + 
We know that 
= 
, so we get
+ 
+ 
+ 
+ $({11^{\log_{11}25})^{1/2}$ (Error compiling LaTeX. Unknown error_msg)
The 
and the cancel out to make 
, and we can do this for the other two terms. We obtain
+ 
+ 
= 
+ 
+ 
= 
.
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 | ||
