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

(New page: == Problem == Suppose that <math>a</math>, <math>b</math>, and <math>c</math> are positive real numbers such that <math>a^{\log_3 7} = 27</math>, <math>b^{\log_7 11} = 49</math>, and <math...)
 
(Solution)
Line 4: Line 4:
  
 
== Solution ==
 
== Solution ==
 +
 +
'''Solution 1'''
  
 
First, we have:
 
First, we have:
Line 50: Line 52:
  
 
and therefore the answer is <math>343+121+5 = \boxed{469}</math>.
 
and therefore the answer is <math>343+121+5 = \boxed{469}</math>.
 +
 +
'''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
 +
 +
<math>a^{(log_a27)(log_37)}</math> + <math>b^{(log_b49)(log_711)</math> + <math>c^{(log_c\sqrt {11})(log_{11}25)}</math>
 +
 +
We know that <math>x^{log_xy}</math> = <math>y</math>, so we get
 +
 +
<math>27^{log_37}</math> + <math>49^{log_711}</math> + <math>\sqrt {11}^{log_{11}25}</math>
 +
 +
<math>(3^{log_37})^3</math> + <math>(7^{log_711})^2</math> + <math>{11^{log_{11}25}^1/2</math>
 +
 +
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
 +
 +
<math>7^3</math> + <math>11^2</math> + <math>25^{1/2}</math>
 +
 +
= <math>343</math> + <math>121</math> + <math>5</math>
 +
= <math>\boxed {469}</math>.
 +
 
== 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}}

Revision as of 13:45, 18 April 2009

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

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 get

$a^{(log_a27)(log_37)}$ + $b^{(log_b49)(log_711)$ (Error compiling LaTeX. ! Missing } inserted.) + $c^{(log_c\sqrt {11})(log_{11}25)}$

We know that $x^{log_xy}$ = $y$, so we get

$27^{log_37}$ + $49^{log_711}$ + $\sqrt {11}^{log_{11}25}$

$(3^{log_37})^3$ + $(7^{log_711})^2$ + ${11^{log_{11}25}^1/2$ (Error compiling LaTeX. ! Double superscript.)

The $3$ and the $log_37$ cancel out to make $7$, and we can do this for the other two terms. We obtain

$7^3$ + $11^2$ + $25^{1/2}$

= $343$ + $121$ + $5$ = $\boxed {469}$.

See Also

2009 AIME II (ProblemsAnswer KeyResources)
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
Invalid username
Login to AoPS