Question

[Bảo Ngọc cute]

Tìm x thuộc Z

a)\(32^{-n}.16^n=2048\)

b)\(2^{-1}.2^n+4.2^n=9.2^5\)

Answer 1

[Phạm Nguyễn Tất Đạt]

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

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

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

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

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

\(2^n=64\)

\(2^n=2^6\)

\(\Rightarrow n=6\)

Answer 2

[Nguyễn Thanh Vân]

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