Difference between revisions of "2007 AMC 10A Problems/Problem 9"

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<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 ==
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== 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>
  
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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>.
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== Solution 2 ==
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Simplify equation 1 which is 3^a=81^b+2, to 3^a=3^4(b+2), which equals 3^a=3^4b+8.
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And
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Simplify equation 2 which is 125^b=5^a-3, to 5^3(b)=5^a-3, which equals 5^3b=5^a-3.
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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.
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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.
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4b+8=3b+3
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b=-5
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Now plug b, which is (-5) back into one of the two earlier equations.
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4(-5)+8=a
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-20+8=a
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a=-12
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(-12)(-5)=60
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Therefore the correct answer is C
  
 
== See also ==
 
== See also ==

Revision as of 11:37, 16 February 2016

Problem

Real numbers $a$ and $b$ satisfy the equations $3^{a} = 81^{b + 2}$ and $125^{b} = 5^{a - 3}$. What is $ab$?

$\text{(A)}\ -60 \qquad \text{(B)}\ -17 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 60$

Solution 1

\[81^{b+2} = 3^{4(b+2)} = 3^a \Longrightarrow a = 4b+8\]

And

\[125^{b} = 5^{3b} = 5^{a-3} \Longrightarrow a - 3 = 3b\]

Substitution gives $4b+8 - 3 = 3b \Longrightarrow b = -5$, and solving for $a$ yields $-12$. Thus $ab = 60\ \mathrm{(E)}$.

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 (ProblemsAnswer KeyResources)
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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