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Difference between revisions of "2009 AIME II Problems/Problem 2"
m (→‎Solution 2)
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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
Line 61: Line 61:
 
<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  

Revision as of 18:19, 9 March 2015

Problem

Suppose that $a$, $b$, and $c$ are positive real numbers such that $a^{\log_3 7} = 27$, $b^{\log_7 11} = 49$, and $c^{\log_{11}25} = \sqrt{11}$. Find \[a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}.\]

Solution 1

First, we have: \[x^{(\log_y z)^2} = x^{\left( (\log_y z)^2 \right) } = x^{(\log_y z) \cdot (\log_y z) } = \left( x^{\log_y z} \right)^{\log_y z}\]

Now, let $x=y^w$, then we have: \[x^{\log_y z} = \left( y^w \right)^{\log_y z} = y^{w\log_y z} = y^{\log_y (z^w)} = z^w\]

This is all we need to evaluate the given formula. Note that in our case we have $27=3^3$, $49=7^2$, and $\sqrt{11}=11^{1/2}$. We can now compute:

\[a^{(\log_3 7)^2} = \left( a^{\log_3 7} \right)^{\log_3 7} = 27^{\log_3 7} = (3^3)^{\log_3 7} = 7^3 = 343\]

Similarly, we get \[b^{(\log_7 11)^2} = (7^2)^{\log_7 11} = 11^2 = 121\]

and \[c^{(\log_{11} 25)^2} = (11^{1/2})^{\log_{11} 25} = 25^{1/2} = 5\]

and therefore the answer is $343+121+5 = \boxed{469}$.

Solution 2

We know from the first three equations that $\log_a27$ = $\log_37$, $\log_b49$ = $\log_711$, and $\log_c\sqrt{11}$ = $\log_{11}25$. Substituting, we get

$a^{(\log_a27)(\log_37)}$ + $b^{(\log_b49)(\log_711)}$ + $c^{(\log_c\sqrt {11})(\log_{11}25)}$

We know that $x^{\log_xy}$ = $y$, so we get

$27^{\log_37}$ + $49^{\log_711}$ + $\sqrt {11}^{\log_{11}25}$

$(3^{\log_37})^3$ + $(7^{\log_711})^2$ + $({11^{\log_{11}25}})^{1/2}$

The $3$ and the $\log_37$ cancel out to make $7$, and we can do this for the other two terms. We obtain

$7^3$ + $11^2$ + $25^{1/2}$

= $343$ + $121$ + $5$ = $\boxed {469}$.

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

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