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 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, $(256)^{.16} \cdot (256)^{.09}=\boxed{\textbf{(A) }\ 4}.$

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1959 AHSME Problems/Problem 5

Problem 5

Solution