1961 AHSME Problems/Problem 9: Difference between revisions
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Let <math>r</math> be the result of doubling both the base and | == Problem 9== | ||
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>. | |||
If <math>r</math> equals the product of <math>a^b</math> by <math>x^b</math>, then <math>x</math> equals: | |||
<math>\textbf{(A)}\ a \qquad | |||
\textbf{(B)}\ 2a \qquad | |||
\textbf{(C)}\ 4a \qquad | |||
\textbf{(D)}\ 2\qquad | |||
\textbf{(E)}\ 4 </math> | |||
==Solution== | |||
From the problem, <math>r = (2a)^{2b}</math>, so | |||
<cmath>(2a)^{2b} = a^b \cdot x^b</cmath> | |||
<cmath>(4a^2)^b = (ax)^b</cmath> | |||
<cmath>4a^2 = ax</cmath> | |||
<cmath>x = 4a</cmath> | |||
Thus, the answer is <math>\boxed{\textbf{(C)}}</math>. | |||
==See Also== | |||
{{AHSME 40p box|year=1961|num-b=8|num-a=10}} | |||
{{MAA Notice}} | |||
[[Category:Introductory Algebra Problems]] | |||
Latest revision as of 13:36, 19 May 2018
Problem 9
Let
be the result of doubling both the base and exponent of
, and
does not equal to
.
If
equals the product of
by
, then
equals:
Solution
From the problem,
, so
Thus, the answer is
.
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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