If $$ 3^{4x} = \left ( 81 \right )^{-1}$$ and $$ 10^{1/y} =0.0001,$$ find the value of $$ 2^{-x+4y}.$$
Correct option is A. 1
Given,
$$3^{4x}=(81)^{-1}$$
$$3^{4x}=(3^4)^{-1}$$
$$3^{4x}=3^{-4}$$
By comparing powers
$$4x=-4$$
$$\therefore x=-1$$
$$10^{\frac{1}{y}}=0.0001$$
$$10^{\frac{1}{y}}=10^{-4}$$
By comparing powers
$$\dfrac{1}{y}=-4$$
$$\therefore y=-\dfrac{1}{4}$$
Now $$2^{-x+4y}$$
$$=2^{-(-1)+4\left ( -\frac{1}{4} \right )}$$
$$=2^{1-1}$$
$$=2^0=1$$