Difference between revisions of "2003 AMC 10B Problems/Problem 9"

(Created page with "==Problem== Find the value of <math>x</math> that satisfies the equation <math>25^{-2} = \frac{5^{48/x}}{5^{26/x} \cdot 25^{17/x}}.</math> <math>\textbf{(A) } 2 \qquad\textbf{(...")
 
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Manipulate the powers of <math>5</math> in order to get a clean expression.
 
Manipulate the powers of <math>5</math> in order to get a clean expression.
  
<math>25^{-2} = \frac{5^{48/x}}{5^{26/x} \cdot 25^{17/x}}</math>
+
<cmath>\frac{5^{\frac{48}{x}}}{5^{\frac{26}{x}} \cdot 25^{\frac{17}{x}}} = \frac{5^{\frac{48}{x}}}{5^{\frac{26}{x}} \cdot 5^{\frac{34}{x}}} = 5^{\frac{48}{x}-(\frac{26}{x}+\frac{34}{x})} = 5^{-\frac{12}{x}}</cmath>
 
+
<cmath>25^{-2} = (5^2)^{-2} = 5^{-4}</cmath>
<math>(5^2)^{-2} = \frac{5^{48/x}}{5^{26/x} \cdot (5^2)^{17/x}}</math>
+
<cmath>5^{-4} = 5^{-\frac{12}{x}}</cmath>
 
 
<math>5^{-4} = \frac{5^{48/x}}{5^{26/x} \cdot 5^{34/x}}</math>
 
 
 
<math>5^{-4} = 5^{48/x-26/x-34/x}</math>
 
 
 
<math>5^{-4} = 5^{(48-26-34)/x}</math>
 
 
 
<math>5^{-4} = 5^{-12/x}</math>
 
  
 
If two numbers are equal, and their bases are equal, then their exponents are equal as well.  Set the two exponents equal to each other.
 
If two numbers are equal, and their bases are equal, then their exponents are equal as well.  Set the two exponents equal to each other.
  
<math>-4=\frac{-12}{x}</math>
+
<cmath>\begin{align*}-4&=\frac{-12}{x}\\
 
+
-4x&=-12\\
<math>-4x=-12</math>
+
x&=\boxed{\mathrm{(B) \ } 3}\end{align*}</cmath>
 
 
<math>\boxed{x=3 \text{ (B)}}</math>
 
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2003|ab=B|num-b=8|num-a=10}}
 
{{AMC10 box|year=2003|ab=B|num-b=8|num-a=10}}

Revision as of 03:35, 10 June 2011

Problem

Find the value of $x$ that satisfies the equation $25^{-2} = \frac{5^{48/x}}{5^{26/x} \cdot 25^{17/x}}.$

$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 9$

Solution

Manipulate the powers of $5$ in order to get a clean expression.

\[\frac{5^{\frac{48}{x}}}{5^{\frac{26}{x}} \cdot 25^{\frac{17}{x}}} = \frac{5^{\frac{48}{x}}}{5^{\frac{26}{x}} \cdot 5^{\frac{34}{x}}} = 5^{\frac{48}{x}-(\frac{26}{x}+\frac{34}{x})} = 5^{-\frac{12}{x}}\] \[25^{-2} = (5^2)^{-2} = 5^{-4}\] \[5^{-4} = 5^{-\frac{12}{x}}\]

If two numbers are equal, and their bases are equal, then their exponents are equal as well. Set the two exponents equal to each other.

\begin{align*}-4&=\frac{-12}{x}\\ -4x&=-12\\ x&=\boxed{\mathrm{(B) \ } 3}\end{align*}

See Also

2003 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions