Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

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

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

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_2&oldid=219192"