1959 AHSME Problems/Problem 5

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 the 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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