Question

Tìm x :

\(\left(\dfrac{1}{9}\right)^x=\left(\dfrac{1}{27}\right)^6\)

Answer 1

\(\left(\dfrac{1}{9}\right)^x=\left(\dfrac{1}{27}\right)^6\)

\(\Rightarrow\left(3^{-2}\right)^x=\left(3^{-3}\right)^6\)

\(\Rightarrow3^{-2x}=3^{-18}\)

\(\Rightarrow-2x=-18\)

\(\Rightarrow x=9\)

Vậy \(x=9\)

Answer 2

\(\left(\dfrac{1}{9}\right)^x=\left(\dfrac{1}{27}\right)^6\)

\(3^{-2x}=3^{-18}\)

=>\(-2x=-18\)

=>\(x=9\)

Vậy...