Theo đề ta có : \(32^{-n}.16^n=2048\)

\(\Rightarrow\frac{1}{32^n}.16^n=2048\)

\(\Rightarrow\frac{16^n}{32^n}=2048\)

\(\Rightarrow\left(\frac{16}{32}\right)^n=\left(\frac{1}{2}\right)^n=2048\)

\(\Rightarrow\frac{1}{2^n}=2048\)

\(\Rightarrow2^n=\frac{1}{2048}\)

\(\Rightarrow2^n=\frac{1}{2^{11}}\Rightarrow1=2^n.2^{11}\)

\(\Rightarrow2^n=2^{-11}\Rightarrow n=-11\) ( bởi vì tích của 2 số nghịch đảo bao giờ cũng bằng 1)

qui ước \(x^{-a}=\frac{1}{x^a}\)

ta có

\(32^{-n}.16^n=2048\Rightarrow\frac{1}{32^n}.16^n=2^{10}\Rightarrow\frac{16^n}{32^n}=2^{10}\)

\(\Rightarrow\left(\frac{16}{32}\right)^n=\frac{1}{2^n}=2^{10}\Rightarrow2^{-n}=2^{10}\Rightarrow-n=10\Rightarrow n=-10\)

a)\(32^{-n}\cdot16^n=2048\)

\(\left(2^5\right)^{-n}\cdot\left(2^4\right)^n\)=2048

\(2^{-5n}\cdot2^{4n}\)=\(2^{11}\)

\(2^{-5n+4n}=2^{11}\)

\(2^{-x}=2^{11}\)

\(\Rightarrow x=-11\)

b)\(2^{-1}\cdot2^n+4\cdot2^n=9\cdot2^5\)

\(\frac{1}{2}\cdot2^n+4\cdot2^n=288\)

\(2^n\left(\frac{1}{2}+4\right)=288\)

\(2^n\cdot\frac{9}{2}=288\)

\(2^n=288:\frac{9}{2}\)

\(2^n=64\)

\(2^n=2^6\)

\(\Rightarrow n=6\)

a) 32^-n . 16ⁿ = 2048

\(\frac{1}{32n}\) . 16ⁿ = 2048

\(\frac{1}{2^n.16^n}\) . 16ⁿ = 2048

\(\frac{1}{2^n}\) = 2048

2^-n = 2048

2^-n = 2¹¹

\(\Rightarrow\) -n = 11

\(\Rightarrow\) n = -11

Vậy n = -11

32^-n.16^n=2048=>1/32^n.16^n=2048

=>1/(16^n.2^n).16^N=2048

=>1/2^n=2048=>n= -11

a) Ta có: \(\frac{1}{9}\cdot27^n=3^n\)

\(\Leftrightarrow\frac{1}{3^2}\cdot\left(3^3\right)^n=3^n\)

\(\Leftrightarrow3^{3n}=3^{n+2}\)

\(\Rightarrow3n=n+2\)

\(\Rightarrow n=1\)

b) Ta có: \(3^2.3^4.3^n=3^7\)

\(\Rightarrow3^n=3\)

\(\Rightarrow n=1\)

c) Ta có: \(2^{-1}.2^n+4.2^n=9.2^5\)

\(\Leftrightarrow2^n\cdot\frac{9}{2}=9.2^5\)

\(\Rightarrow2^n=2^6\)

\(\Rightarrow n=6\)

d) Ta có: \(32^{-n}.16^n=2048\)

\(\Leftrightarrow\frac{1}{2^{5n}}\cdot2^{4n}=2^{11}\)

\(\Leftrightarrow2^{4n}=2^{5n+11}\)

\(\Rightarrow4n=5n+11\)

\(\Rightarrow n=-11\)

32ⁿ - 16ⁿ = 2048

=> (2⁵)ⁿ - (2⁴)ⁿ = 2¹¹

=> 2ⁿ(2⁵ - 2⁴) = 2¹¹

=> 2ⁿ.2³ = 2¹¹

=> 2ⁿ = 2⁸

a) \(\frac{1}{9}.27^n=3^n\)

\(\Leftrightarrow3^{-2}.3^{3n}=3^n\)

\(\Leftrightarrow3^{3n-2}=3^n\)

\(\Leftrightarrow3n-2=n\)

\(\Leftrightarrow2n=2\)

\(\Leftrightarrow n=1\)

b)\(3^{-2}.3^4.3^n=3^7\)

\(\Leftrightarrow3^{2+n}=3^7\)

\(\Leftrightarrow2+n=7\)

\(\Leftrightarrow n=5\)

c) \(2^{-1}.2^n+4.2^n=9.2^5\)

\(\Leftrightarrow2^n\left(4+\frac{1}{2}\right)=9.2^5\)

\(\Leftrightarrow2^n.\frac{9}{2}=9.2^5\)

\(\Leftrightarrow2^{n-1}=2^5\)

\(\Leftrightarrow n-1=5\Leftrightarrow n=6\)