Difference between revisions of "2016 AMC 10A Problems/Problem 2"
(Added some elements from old 12A solution) |
|||
Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
− | We can rewrite <math>10^{x}\cdot 100^{2x}=1000^{5}</math> as <math>10^{5x}=10^{15}</math> | + | We can rewrite <math>10^{x}\cdot 100^{2x}=1000^{5}</math> as <math>10^{5x}=10^{15}</math>: |
− | + | <cmath>10^x\cdot100^{2x}=10^x\cdot(10^2)^{2x}</cmath> | |
+ | <cmath>10^x\cdot10^{4x}=(10^3)^5</cmath> | ||
+ | <cmath>10^{5x}=10^{15}</cmath> | ||
+ | Since the bases are equal, we can set the exponents equal: <math>5x=15</math>. Solving gives us: <math>x = \boxed{\textbf{(C)}\;3.}</math> | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=A|num-b=1|num-a=3}} | {{AMC10 box|year=2016|ab=A|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:46, 4 February 2016
Problem
For what value of does ?
Solution
We can rewrite as : Since the bases are equal, we can set the exponents equal: . Solving gives us:
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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.