ta có công thức như sau :

\(a^{-x}=?\)

lời giải công thức này như sau :

\(a^{-x}=\left(\frac{1}{a}\right)^x\)

vậy bài cũng gải tương tự 

\(32^{-x}.16^x=\left(\frac{1}{32}\right)^x.\left(16^x\right)\)

\(=\left(\frac{16}{32}\right)^x=\left(\frac{1}{2}\right)^x=2^{-x}\)

mà \(2048=2^{11}\)

 \(\Rightarrow-x=11\)

 \(\Leftrightarrow x=-11\)

vậy \(x=-11\)

\(\Rightarrow\)\(\left(\frac{1}{32}\right)^x\cdot16^x=2048\)

\(\Rightarrow\)\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^{-11}\)

\(\Rightarrow\)\(x=-11\)

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

x = -11 do ban

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