AIME 2009 II · 第 2 题

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

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

Solution 3

First, let us take the log base 3 of the first expression. We get $\log_3{a^{\log_3{7}}} = 3$. Simplifying, we get

$$(\log_3{7})(\log_3{a}) = 3$$ . So,

$$\log_3 a = \frac{3}{\log_3{7}}$$ , and

$$a = 3^\frac{3}{log_3{7}}$$ . We can repeat the same process for the other equations, giving us

$$b = 7^\frac{2}{\log_7{11}}$$ , and

$$c = (\sqrt{11})^\frac{1}{\log_ {11}{25}}$$ . Raising $a$ to the power of $(\log_3{7})^2$, we get

$$3^{3\log_3{7}} = 3^{\log_3{343}} = 343$$ . Repeating a similar process for the other ones(you have to turn the square root to a fractional power for c), we get $343+121+5 = \boxed{469}$

~idk12345678