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Difference between revisions of "2009 AIME II Problems/Problem 2"
 
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== Solution 2 ==
 
== Solution 2 ==
  
−
We know from the first three equations that <math>\log_a27</math> = <math>\log_37</math>, <math>\log_b49</math> = <math>\log_711</math>, and <math>\log_c\sqrt{11}</math> = <math>\log_{11}25</math>. Substituting, we get
+
We know from the first three equations that <math>\log_a27 = \log_37</math>, <math>\log_b49 = \log_711</math>, and <math>\log_c\sqrt{11} = \log_{11}25</math>. Substituting, we find
  
−
<math>a^{(\log_a27)(\log_37)}</math> + <math>b^{(\log_b49)(\log_711)</math> + <math>c^{(\log_c\sqrt {11})(\log_{11}25)}</math>
+
<cmath>a^{(\log_a27)(\log_37)} + b^{(\log_b49)(\log_711)} + c^{(\log_c\sqrt {11})(\log_{11}25)}.</cmath>
  
−
We know that <math>x^{\log_xy}</math> = <math>y</math>, so we get
+
We know that <math>x^{\log_xy} =y</math>, so we find
  
−
<math>27^{\log_37}</math> + <math>49^{\log_711}</math> + <math>\sqrt {11}^{\log_{11}25}</math>
+
<cmath>27^{\log_37} + 49^{\log_711} + \sqrt {11}^{\log_{11}25}</cmath>
  
−
<math>(3^{\log_37})^3</math> + <math>(7^{\log_711})^2</math> + <math>({11^{\log_{11}25})^{1/2}</math>
+
<cmath>(3^{\log_37})^3 + (7^{\log_711})^2 + ({11^{\log_{11}25}})^{1/2}.</cmath>
  
−
The <math>3</math> and the <math>\log_37</math> cancel out to make <math>7</math>, and we can do this for the other two terms. We obtain
+
The <math>3</math> and the <math>\log_37</math> cancel to make <math>7</math>, and we can do this for the other two terms. Thus, our answer is
  
−
<math>7^3</math> + <math>11^2</math> + <math>25^{1/2}</math>
+
<cmath>7^3 + 11^2 + 25^{1/2}</cmath>
 +
<cmath>= 343 + 121 + 5</cmath>
 +
<cmath>= \boxed {469}.</cmath>
  
−
= <math>343</math> + <math>121</math> + <math>5</math>
+
== Solution 3 ==
−
= <math>\boxed {469}</math>.
+
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 ==
 
== See Also ==
  
 
{{AIME box|year=2009|n=II|num-b=1|num-a=3}}
 
{{AIME box|year=2009|n=II|num-b=1|num-a=3}}
 +
[[Category: Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

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

See Also

2009 AIME II (Problems • Answer Key • Resources)
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Problem 1 		Followed by
Problem 3
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All AIME Problems and Solutions

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