# 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 out to make $7$, and we can do this for the other two terms. We obtain $7^3 + 11^2 + 25^{1/2}$ $= 343 + 121 + 5$ $= \boxed {469}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 