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Difference between revisions of "1959 AHSME Problems/Problem 5"

(Created page with "== Problem 5== The value of <math>\left(256\right)^{.16}\left(256\right)^{.09}</math> is: <math>\textbf{(A)}\ 4 \qquad \\ \textbf{(B)}\ 16\qquad \\ \textbf{(C)}\ 64\qquad \\ ...")
 
(Added See Also Section)
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Since <math>256 =16^2=(4^2)^2=4^4</math>, we can solve for the fourth root of <math>256</math>: <cmath>\sqrt[4]{256}=\sqrt[4]{4^4}=4.</cmath>
 
Since <math>256 =16^2=(4^2)^2=4^4</math>, we can solve for the fourth root of <math>256</math>: <cmath>\sqrt[4]{256}=\sqrt[4]{4^4}=4.</cmath>
 
Therefore, <math>(256)^{.16} \cdot (256)^{.09}=\boxed{\textbf{(A) }\ 4}.</math>
 
Therefore, <math>(256)^{.16} \cdot (256)^{.09}=\boxed{\textbf{(A) }\ 4}.</math>
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== See also ==
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{{AHSME 50p box|year=1959|num-b=4|num-a=6}}
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{{MAA Notice}}

Revision as of 20:57, 25 February 2018

Problem 5

The value of $\left(256\right)^{.16}\left(256\right)^{.09}$ is:

$\textbf{(A)}\ 4 \qquad \\ \textbf{(B)}\ 16\qquad \\ \textbf{(C)}\ 64\qquad \\ \textbf{(D)}\ 256.25\qquad \\ \textbf{(E)}\ -16$

Solution

When we multiply numbers with exponents, we add tne exponents together and leave the bases unchanged. We can apply this concept to computate $256^{0.16} \cdot 256^{0.09}$: \[256^{0.16} \cdot 256^{0.09} = 256^{0.16+0.09}=256^{0.25}.\] Now we can convert the decimal exponent to a fraction: \[256^{0.25} = 256^{\frac{1}{4}}.\] Now, let us convert the expression into radical form. Since $4$ is the denominator of the fractional exponent, it will be the index exponent: \[256^{\frac{1}{4}}=\sqrt[4]{256}.\] Since $256 =16^2=(4^2)^2=4^4$, we can solve for the fourth root of $256$: \[\sqrt[4]{256}=\sqrt[4]{4^4}=4.\] Therefore, $(256)^{.16} \cdot (256)^{.09}=\boxed{\textbf{(A) }\ 4}.$

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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All AHSME Problems and Solutions

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