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

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<math>4b+8=3b+3 \Longrightarrow b=-5</math>
 
<math>4b+8=3b+3 \Longrightarrow b=-5</math>
  
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Now plug <math>b</math>, which is <math>(-5)</math> back into one of the two earlier equations.
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Now plug <math>b</math>, which is <math>-5</math>, back into one of the two earlier equations.
 
<math>4(-5)+8=a</math>
 
<math>4(-5)+8=a</math>
 
<math>-20+8=a</math>
 
<math>-20+8=a</math>

Revision as of 11:38, 3 June 2021

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 1 another similar way

Simplify equation $1$, which is $3^a=81^{b+2}$, to $3^a=3^{4b+8}$.

And

Simplify equation $2$, which is $125^b=5^{a-3}$, to $5^{3b}=5^{a-3}$.

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 the values for $a$ and $b$ are constant for both equations, set the equations equal to each other. $4b+8=3b+3 \Longrightarrow b=-5$

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 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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