Solve for X: x^5/16 = 2 Fifth-Power Equation

Let's solve the given equation:

$\frac{x^5}{16}=2$

First, let's deal with the fact that we need to eliminate the denominator on the left side of the given equation, we'll do this by multiplying both sides of the equation by the common denominator - which is the number 16:

$\frac{x^5}{16}=2 \hspace{8pt}\text{/}\cdot 16\\ x^5=2\cdot 16\\ x^5=32$

From here, we'll solve simply, we'll perform on both sides the opposite operation to the fifth power that applies to the unknown in the equation, which is the fifth root operation, using the laws of exponents:

a. Definition of root as a power:

$\sqrt[n]{a}=a^{\frac{1}{n}}$and two laws of exponents:

b. Law of exponents for power of a power:

$(a^m)^n=a^{m\cdot n}$

Let's continue solving the equation:
$x^5=32 \hspace{8pt}\text{/}\sqrt[5]{\hspace{6pt}}\\ \sqrt[5]{ x^5}=\sqrt[5]{ 32}\\ (x^5)^{\frac{1}{5}}=2\\ x^{5\cdot\frac{1}{5}}=2\\ \boxed{x=2}$

In the first stage, we applied the fifth root to both sides of the equation, then we remembered the definition of root as a power (a.) on the left side, in the next stage - we applied the law of exponents for power of a power (b.) on the left side, and remembered that raising a number to the power of 1 doesn't change the number.

Additionally, we remembered that since an odd-order power preserves the sign of the number it's applied to, taking an odd-order root requires consideration of only one case regarding the sign of the number being rooted (this is unlike taking an even-order root, which requires consideration of two possible cases - positive and negative),

Let's summarize the solution of the equation:

$\frac{x^5}{16}=2 \hspace{8pt}\text{/}\cdot 16\\ x^5=32 \hspace{8pt}\text{/}\sqrt[5]{\hspace{6pt}}\\ \boxed{x=2}$

Therefore, the correct answer is answer B.