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 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

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

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

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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