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

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== Solution 2 ==
 
== Solution 2 ==
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Simplify equation <math>1</math> which is <math>3^a=81^b+2</math>, to <math>3^a=3^4(b+2)</math>, which equals <math>3^a=3^4b+8</math>.  
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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>5^3(b)=5^a-3</math>, which equals <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>.

Revision as of 11:26, 21 January 2018

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 $3^a=3^{4b+8}$.

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 $a$ 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 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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