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 <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>. | ||
| − | Since both equations are equal to <math>a</math>, and <math>a</math> and <math>b</math> are | + | Since both equations are equal to <math>a</math>, and the values for <math>a</math> and <math>b</math> are constant for both equations, set the equations equal to each other. |
| − | <math>4b+8=3b+3 | + | <math>4b+8=3b+3 \Longrightarrow b=-5</math> |
| − | |||
| − | Now plug <math>b</math>, which is <math> | + | Now plug <math>b</math>, which is <math>-5</math>, back into one of the two earlier equations. |
| − | <math>4(-5)+8=a | + | <math>4(-5)+8=a \Longrightarrow -20+8=a \Longrightarrow a=-12</math> |
| − | |||
| − | |||
<math>(-12)(-5)=60</math> | <math>(-12)(-5)=60</math> | ||
| − | Therefore the correct answer is E | + | Therefore the correct answer is <math>\mathrm{(E)}</math> |
== See also == | == See also == | ||
Latest revision as of 11:40, 3 June 2021
Problem
Real numbers 
and 
satisfy the equations 
and 
. What is 
?
Solution 1
![\[81^{b+2} = 3^{4(b+2)} = 3^a \Longrightarrow a = 4b+8\]](http://latex.artofproblemsolving.com/0/f/2/0f2765a3b43aef194ebe11a4a63c5083344208ce.png)
And
![\[125^{b} = 5^{3b} = 5^{a-3} \Longrightarrow a - 3 = 3b\]](http://latex.artofproblemsolving.com/d/e/c/decae380961b77b05f8b9876d54f4c9777d2e402.png)
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 the values for and are constant for both equations, set the equations equal to each other.
Now plug , which is 
, back into one of the two earlier equations.
Therefore the correct answer is 
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. 
