Assertion (A): If $\log_x \space \dfrac{1}{8} = -\dfrac{1}{3}$, then the value of x is 2.

Reason (R): If n^x = m, then log_n m = x.

1. A is true, R is false

2. A is false, R is true

3. Both A and R are true

4. Both A and R are false

Given,

$\Rightarrow \log_x \space \dfrac{1}{8} = -\dfrac{1}{3} \\[1em] \Rightarrow x^{-\dfrac{1}{3}} = \dfrac{1}{8} \\[1em] \Rightarrow \Big(\dfrac{1}{x}\Big)^{\dfrac{1}{3}} = \Big(\dfrac{1}{8}\Big) \\[1em] \text{ Cubing on both sides, we get:} \\[1em] \Rightarrow \Big[\Big(\dfrac{1}{x}\Big)^{\dfrac{1}{3}}\Big]^3 = \Big(\dfrac{1}{8}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{1}{x}\Big)^{\dfrac{1}{3} \times 3} = \Big(\dfrac{1}{512}\Big) \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{1}{512} \\[1em] \Rightarrow x = 512.$

So, Assertion (A) is false.

If n^x = m, then log_n m = x.

This is the fundamental definition of a logarithm.

So, Reason (R) is true.

A is false, R is true.

Hence, option 2 is the correct option.