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{(...") |
|||
(2 intermediate revisions by one other user not shown) | |||
Line 9: | Line 9: | ||
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. | ||
− | < | + | <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> | |
− | + | <cmath>5^{-4} = 5^{-\frac{12}{x}}</cmath> | |
− | |||
− | |||
− | |||
− | |||
− | |||
− | < | ||
− | |||
− | < | ||
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. | ||
− | < | + | <cmath>\begin{align*}-4&=\frac{-12}{x}\\ |
− | + | -4x&=-12\\ | |
− | + | 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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 11:09, 4 July 2013
Problem
Find the value of that satisfies the equation
Solution
Manipulate the powers of in order to get a clean expression.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.