Difference between revisions of "2003 AMC 10B Problems/Problem 9"
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<cmath>\begin{align*}-4&=\frac{-12}{x}\\ | <cmath>\begin{align*}-4&=\frac{-12}{x}\\ | ||
-4x&=-12\\ | -4x&=-12\\ | ||
| − | x&=\boxed{\ | + | x&=\boxed{\textbf{(B) \ } 3}\end{align*}</cmath> |
==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 19:01, 26 November 2011
Problem
Find the value of 
that satisfies the equation
Solution
Manipulate the powers of 
in order to get a clean expression.
![\[\frac{5^{\frac{48}{x}}}{5^{\frac{26}{x}} \cdot 25^{\frac{17}{x}}} = \frac{5^{\frac{48}{x}}}{5^{\frac{26}{x}} \cdot 5^{\frac{34}{x}}} = 5^{\frac{48}{x}-(\frac{26}{x}+\frac{34}{x})} = 5^{-\frac{12}{x}}\]](http://latex.artofproblemsolving.com/e/e/7/ee7b66e25c28807e2bfea090fc8f89c5d6b41aa2.png)
![\[25^{-2} = (5^2)^{-2} = 5^{-4}\]](http://latex.artofproblemsolving.com/5/c/d/5cdb1a2ed954ee8fda8aae01c8a9032b16e28bba.png)
![\[5^{-4} = 5^{-\frac{12}{x}}\]](http://latex.artofproblemsolving.com/2/1/9/219e5f75112057ae2f3f02acf33d2ea76bc8581b.png)
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.
See Also
| 2003 AMC 10B (Problems • Answer Key • Resources) | ||
| 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 | ||
