If $9 \times 81^x = \dfrac{1}{27^{(x - 3)}}$, then x =

1. 0

2. -1

3. 1

4. 3

Given,

$\Rightarrow 9 \times 81^x = \dfrac{1}{27^{x - 3}} \\[1em] \Rightarrow 3^2 \times (3^4)^{x} = \dfrac{1}{3^{3(x - 3)}} \\[1em] \Rightarrow 3^2 \times 3^{4x} = 3^{-3(x - 3)} \\[1em] \Rightarrow 3^{2 + 4x} = 3^{-3 \times x - 3 \times - 3} \\[1em] \Rightarrow 3^{2 + 4x} = 3^{-3x + 9}$

Equating the exponents:

$\Rightarrow 2 + 4x = -3x + 9 \\[1em] \Rightarrow 4x + 3x = 9 - 2 \\[1em] \Rightarrow 7x = 7 \\[1em] \Rightarrow x = \dfrac{7}{7} \\[1em] \Rightarrow x = 1.$

Hence, option 3 is the correct option.