Question

(2/3)^abslx+2l=(81/16)^-3

Answer 1

\[(\frac{81}{16})^{-3}=(\frac{16}{81})^3=(\frac{2^4}{3^4})^3\]
\[=(\frac{2}{3})^{12}\]

Answer 2

this tells you
\[|x+2|=12\] so
\[x=2=12\] or
\[x+2=-12\]