If 3^4x = 81^-1 and $(10)^{\dfrac{1}{y}}$ = 0.0001, find the value of 2^-x.(16)^y.

Given,

⇒ 3^4x = 81^-1

⇒ 3^4x = (3⁴)^-1

⇒ 3^4x = (3^-4)

⇒ 4x = -4

⇒ x = -1.

Given,

$\Rightarrow (10)^{\dfrac{1}{y}} = 0.0001 \\[1em] \Rightarrow (10)^{\dfrac{1}{y}} = \dfrac{1}{10^4} \\[1em] \Rightarrow (10)^{\dfrac{1}{y}} = 10^{-4} \\[1em] \Rightarrow \dfrac{1}{y} = -4 \\[1em] \Rightarrow y = -\dfrac{1}{4}.$

Substituting value of x and y in 2^-x.(16)^y we get,

$\Rightarrow 2^{-1}.(16)^{-\dfrac{1}{4}} \\[1em] \Rightarrow 2 \times (2^4)^{-\dfrac{1}{4}} \\[1em] \Rightarrow 2 \times 2^{4 \times -\dfrac{1}{4}} \\[1em] \Rightarrow 2 \times 2^{-1} \\[1em] \Rightarrow 2 \times \dfrac{1}{2} \\[1em] \Rightarrow 1.$

Hence, 2^-x.(16)^y = 1.