Basic expo

$81^{4^x}$ is equal to $9^{2(4^x)}$ , or more simplified form, $9^{2^{2x+1}}$

$9^{2^{2x+1}} = 9 \rightarrow 2^{2x+1}=1 \rightarrow 2x+1 = 0 \rightarrow x = -\frac{1}{2}$

$81^{4^x}=9=\sqrt{81}=81^{\frac12}$ $\Rightarrow 4^x=\frac12$ $\Rightarrow 4^x=\frac{1}{\sqrt4}=\frac{1}{4^{\frac12}}=4^{-\frac12}$ $\therefore x=-\frac12=\color{#0C6AC7} {\boxed {-0.5}}$

You know that 81^0.5 = 9 you need to solve 4^x = 0.5

since 4^{0.5} = 2 and the reciporical of 2 = 1/2 Meaning 4^{-1/2} = 0.5

81^{0.5} = 9

x = -0.5

Done the same