Question

Answer 1

1)\(32^{-n}.16^n=\dfrac{1}{1024}\Leftrightarrow2^{-5n}.2^{4n}=\dfrac{1}{2^{10}}\Leftrightarrow2^{-n}=\dfrac{1}{2^{10}}\Leftrightarrow2^n=2^{10}\Leftrightarrow n=10\)

Answer 2

1: Ta có: \(32^{-n}\cdot16^n=\dfrac{1}{1024}\)

\(\Leftrightarrow\dfrac{1}{32^n}\cdot16^n=\dfrac{1}{1024}\)

\(\Leftrightarrow2^n=1024\)

hay n=10