Difference between revisions of "2009 AIME II Problems/Problem 2"

Latest revision as of 14:20, 10 May 2024

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

@@ Line 68: / Line 68: @@
 <cmath>= 343 + 121 + 5</cmath>
 <cmath>= \boxed {469}.</cmath>
+== Solution 3 ==
+First, let us take the log base 3 of the first expression. We get <math>\log_3{a^{\log_3{7}}} = 3</math>. Simplifying, we get <cmath>(\log_3{7})(\log_3{a}) = 3</cmath>. So, <cmath>\log_3 a = \frac{3}{\log_3{7}}</cmath>, and <cmath>a = 3^\frac{3}{log_3{7}}</cmath>. We can repeat the same process for the other equations, giving us <cmath>b = 7^\frac{2}{\log_7{11}}</cmath>, and <cmath>c = (\sqrt{11})^\frac{1}{\log_
+{11}{25}}</cmath>. Raising <math>a</math> to the power of <math>(\log_3{7})^2</math>, we get <cmath>3^{3\log_3{7}} = 3^{\log_3{343}} = 343</cmath>. Repeating a similar process for the other ones(you have to turn the square root to a fractional power for c), we get <math>343+121+5 = \boxed{469}</math>
+~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
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