The value of x such that $\Big(\dfrac{3}{7}\Big)^3 \times \Big(\dfrac{3}{7}\Big)^{-8} = \Big(\dfrac{3}{7}\Big)^{2x + 3}$ is

1. -4
2. -2
3. 0
4. 1

Given:

$\Big(\dfrac{3}{7}\Big)^3 \times \Big(\dfrac{3}{7}\Big)^{-8} = \Big(\dfrac{3}{7}\Big)^{2x + 3}$

LHS = $\Big(\dfrac{3}{7}\Big)^3 \times \Big(\dfrac{3}{7}\Big)^{-8}$

$= \dfrac{3^3}{7^3} \times \dfrac{3^{-8}}{7^{-8}} \\[1em] = \dfrac{3^3 \times 3^{-8}}{7^3 \times 7^{-8}} \\[1em] = \dfrac{3^{3+(-8)}}{7^{3+(-8)}} = \dfrac{3^{3-8}}{7^{3-8}} \\[1em] = \dfrac{3^{-5}}{7^{-5}} \\[1em] = \Big(\dfrac{3}{7}\Big)^{-5} \\[1em]$

Now, LHS = $\Big(\dfrac{3}{7}\Big)^{-5}$

As the base of both LHS and RHS is same, let us compare the exponents:

-5 = 2x + 3
2x = -5 - 3
2x = -(5 + 3)
2x = -8
x = $\dfrac{-8}{2}$
x = -4

Hence, option 1 is the correct option.