Given that $\sqrt{125^x} = \dfrac{5^{1-x}}{25}$, find the value of $\sqrt{125^x}$. Show that $\log_2 x - \log_{16} x = \dfrac{3\lg x}{4\lg 2}$.

1. Express both sides of the equation as powers of the same base (base 5).

LHS: $\sqrt{125^x} = (5^3)^{\frac{x}{2}} = 5^{\frac{3x}{2}}$

RHS: $\dfrac{5^{1-x}}{25} = \dfrac{5^{1-x}}{5^2} = 5^{1-x-2} = 5^{-1-x}$

2. Equate the indices and solve for $x$.

$5^{\frac{3x}{2}} = 5^{-1-x}$

$\dfrac{3x}{2} = -1 - x$

$3x = -2 - 2x$

$5x = -2 \implies x = -\dfrac{2}{5}$

3. Substitute the value of $x$ back into the expression to find its value.

$\sqrt{125^x} = 5^{\frac{3x}{2}}$

Substitute $x = -\dfrac{2}{5}$:

$5^{\frac{3}{2} \left(-\frac{2}{5}\right)} = 5^{-\frac{3}{5}}$

1. Change the base of the logarithms on the left-hand side to base 10.

$\text{LHS} = \log_2 x - \log_{16} x$
$= \frac{\lg x}{\lg 2} - \frac{\lg x}{\lg 16}$

2. Express $\lg 16$ in terms of $\lg 2$.

$\lg 16 = \lg (2^4) = 4\lg 2$
$= \frac{\lg x}{\lg 2} - \frac{\lg x}{4\lg 2}$

3. Combine the fractions to obtain the right-hand side.

$= \frac{4\lg x}{4\lg 2} - \frac{\lg x}{4\lg 2}$
$= \frac{4\lg x - \lg x}{4\lg 2}$
$= \frac{3\lg x}{4\lg 2} = \text{RHS}$