Difference between revisions of "2007 AMC 10A Problems/Problem 9"
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Substitution gives <math>4b+8 - 3 = 3b \Longrightarrow b = -5</math>, and solving for <math>a</math> yields <math>-12</math>. Thus <math>ab = 60\ \mathrm{(E)}</math>. | Substitution gives <math>4b+8 - 3 = 3b \Longrightarrow b = -5</math>, and solving for <math>a</math> yields <math>-12</math>. Thus <math>ab = 60\ \mathrm{(E)}</math>. | ||
− | == Solution | + | == Solution 1 another similar way == |
− | Simplify equation 1 which is 3^a=81^b+2, to 3^a=3^ | + | 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 2 which is 125^b=5^a-3, to 5^ | + | 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 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 | + | Therefore the correct answer is E |
== See also == | == See also == |
Revision as of 21:58, 17 January 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 |
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