Difference between revisions of "2007 AMC 10A Problems/Problem 9"
(→Solution) |
|||
| Line 4: | Line 4: | ||
<math>\text{(A)}\ -60 \qquad \text{(B)}\ -17 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 60</math> | <math>\text{(A)}\ -60 \qquad \text{(B)}\ -17 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 60</math> | ||
| − | == Solution == | + | == Solution 1 == |
<cmath>81^{b+2} = 3^{4(b+2)} = 3^a \Longrightarrow a = 4b+8</cmath> | <cmath>81^{b+2} = 3^{4(b+2)} = 3^a \Longrightarrow a = 4b+8</cmath> | ||
| Line 12: | Line 12: | ||
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 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. | ||
| + | |||
| + | 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. | ||
| + | |||
| + | 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. | ||
| + | |||
| + | 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. | ||
| + | 4b+8=3b+3 | ||
| + | b=-5 | ||
| + | |||
| + | Now plug b, which is (-5) back into one of the two earlier equations. | ||
| + | 4(-5)+8=a | ||
| + | -20+8=a | ||
| + | a=-12 | ||
| + | |||
| + | (-12)(-5)=60 | ||
| + | |||
| + | Therefore the correct answer is C | ||
== See also == | == See also == | ||
Revision as of 10:37, 16 February 2016
Contents
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 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.
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.
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.
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. 4b+8=3b+3 b=-5
Now plug b, which is (-5) back into one of the two earlier equations. 4(-5)+8=a -20+8=a a=-12
(-12)(-5)=60
Therefore the correct answer is C
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. 
