Difference between revisions of "2007 AMC 10A Problems/Problem 9"
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== Solution 1 another similar way == | == Solution 1 another similar way == | ||
− | Simplify equation <math>1</math> which is <math>3^a=81^{b+2}</math>, to <math>3^a=3^{4b+8}</math>. | + | Simplify equation <math>1</math>, which is <math>3^a=81^{b+2}</math>, to <math>3^a=3^{4b+8}</math>. |
And | And | ||
− | Simplify equation <math>2</math> which is <math>125^b=5^{a-3}</math>, to <math>5^{3b}=5^{a-3}</math>. | + | Simplify equation <math>2</math>, which is <math>125^b=5^{a-3}</math>, to <math>5^{3b}=5^{a-3}</math>. |
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>. | 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>. |
Revision as of 11:35, 3 June 2021
Problem
Real numbers and satisfy the equations and . What is ?
Solution 1
And
Substitution gives , and solving for yields . Thus .
Solution 1 another similar way
Simplify equation , which is , to .
And
Simplify equation , which is , to .
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.