Difference between revisions of "2007 AMC 10A Problems/Problem 9"
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== Solution 2 == | == Solution 2 == | ||
− | Simplify equation 1 which is 3^a=81^b+2, to 3^a=3^4(b+2), which equals 3^a=3^4b+8. | + | Simplify equation <math>1</math> which is <math>3^a=81^b+2</math>, to <math>3^a=3^4(b+2)</math>, which equals <math>3^a=3^4b+8</math>. |
And | And | ||
− | Simplify equation 2 which is 125^b=5^a-3, to 5^3(b)=5^a-3, which equals 5^3b=5^a-3. | + | Simplify equation <math>2</math> which is <math>125^b=5^a-3</math>, to <math>5^3(b)=5^a-3</math>, which equals <math>5^3b=5^a-3</math>. |
− | Now, eliminate the bases from the simplified equations 1 and 2 to arrive at a=4b+8 and 3b=a-3. Rewrite equation 2 so that it is in terms of a. That would be a=3b+3. | + | Now, eliminate the bases from the simplified equations <math>1</math> and <math>2</math> to arrive at <math>a=4b+8</math> and <math>3b=a-3</math>. Rewrite equation <math>2</math> so that it is in terms of <math>a</math>. That would be <math>a=3b+3</math>. |
− | Since both equations are equal to a, and a and b are the same number for both problems, set the equations equal to each other. | + | Since both equations are equal to <math>a</math>, and <math>a</math> and <math>b</math> are the same number for both problems, set the equations equal to each other. |
− | 4b+8=3b+3 | + | <math>4b+8=3b+3</math> |
− | b=-5 | + | <math>b=-5</math> |
− | Now plug b, which is (-5) back into one of the two earlier equations. | + | Now plug <math>b</math>, which is <math>(-5)</math> back into one of the two earlier equations. |
− | 4(-5)+8=a | + | <math>4(-5)+8=a</math> |
− | -20+8=a | + | <math>-20+8=a</math> |
− | a=-12 | + | <math>a=-12</math> |
− | (-12)(-5)=60 | + | <math>(-12)(-5)=60</math> |
Therefore the correct answer is E | Therefore the correct answer is E |
Revision as of 12:02, 23 July 2017
Contents
Problem
Real numbers and satisfy the equations and . What is ?
Solution 1
And
Substitution gives , and solving for yields . Thus .
Solution 2
Simplify equation which is , to , which equals .
And
Simplify equation which is , to , which equals .
Now, eliminate the bases from the simplified equations and to arrive at and . Rewrite equation so that it is in terms of . That would be .
Since both equations are equal to , and and are the same number for both problems, set the equations equal to each other.
Now plug , which is back into one of the two earlier equations.
Therefore the correct answer is E
See also
2007 AMC 10A (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.