2009 AIME II Problems/Problem 2

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 find

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

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

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

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

The $3$ and the $\log_37$ cancel to make $7$ , and we can do this for the other two terms. Thus, our answer is

$7^3 + 11^2 + 25^{1/2}$ $= 343 + 121 + 5$ $= \boxed {469}.$

2009 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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2009 AIME II Problems/Problem 2

Contents

Problem

Solution 1

Solution 2

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