Given: $(5x)^{\frac{1}{8}} = (4x)^{\frac{1}{6}}$

Let's dispense of these fractional exponents by raising both sides of the equation to the power of 24 (24 is the least common multiple of 8 and 6).

We get: $[(5x)^{\frac{1}{8}}]^{24} = [(4x)^{\frac{1}{6}}]^{24}$

Apply the Power of a Power law to get: $(5x)^{3} = (4x)^{4}$

Apply the Power of a Product law to get: $(5^3)(x^3) = (4^4)(x^4)$

Evaluate: $125x^3 = 256x^4$

Divide both sides of the equation by $x^3$ to get: $125 = 256x$

Divide both sides of the equation by $256$ to get: $\frac{125}{256} = x$

So, $x = \frac{125}{256}$ is ONE possible solution to the given equation.

At this point we need to recognize that $x = 0$ is another possible solution since $0^{\frac{1}{8}} = 0^{\frac{1}{6}}$

So the range of all possible values of $x = \frac{125}{256} - 0 = \frac{125}{256}$

So we have:
QUANTITY A: $\frac{125}{256}$
QUANTITY B: $0.5$

No need to grab the calculator.
Since $\frac{125}{250}=0.5$, we can conclude that $\frac{125}{256}$ is less than $0.5$

Answer: B

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