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Difference between revisions of "1961 AHSME Problems/Problem 9"
 
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Let <math>r</math> be the result of doubling both the base and the exponent of <math>a^b</math>, <math>b\not=0</math>. If <math>r</math> equals the product of <math>a^b</math> by <math>x^b</math>, then <math>x</math> equals:
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== Problem 9==
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Let <math>r</math> be the result of doubling both the base and exponent of <math>a^b</math>, and <math>b</math> does not equal to <math>0</math>.  
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If <math>r</math> equals the product of <math>a^b</math> by <math>x^b</math>, then <math>x</math> equals:
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<math>\textbf{(A)}\ a \qquad
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\textbf{(B)}\ 2a \qquad
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\textbf{(C)}\ 4a \qquad
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\textbf{(D)}\ 2\qquad
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\textbf{(E)}\ 4  </math> 
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==Solution==
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From the problem, <math>r = (2a)^{2b}</math>, so
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<cmath>(2a)^{2b} = a^b \cdot x^b</cmath>
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<cmath>(4a^2)^b = (ax)^b</cmath>
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<cmath>4a^2 = ax</cmath>
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<cmath>x = 4a</cmath>
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Thus, the answer is <math>\boxed{\textbf{(C)}}</math>.
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==See Also==
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{{AHSME 40p box|year=1961|num-b=8|num-a=10}}
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{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 14:36, 19 May 2018

Problem 9

Let $r$ be the result of doubling both the base and exponent of $a^b$, and $b$ does not equal to $0$. If $r$ equals the product of $a^b$ by $x^b$, then $x$ equals:

$\textbf{(A)}\ a \qquad \textbf{(B)}\ 2a \qquad \textbf{(C)}\ 4a \qquad \textbf{(D)}\ 2\qquad \textbf{(E)}\ 4$

Solution

From the problem, $r = (2a)^{2b}$, so \[(2a)^{2b} = a^b \cdot x^b\] \[(4a^2)^b = (ax)^b\] \[4a^2 = ax\] \[x = 4a\] Thus, the answer is $\boxed{\textbf{(C)}}$.

See Also

1961 AHSC (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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions


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