Question

Bài 1: tìmn\(\in N\)

d) \(\frac{1}{32^n}.256^n=2048:2^2\)

giúp với mai mk nộp rồi cảm ơn nhiều

Answer 1

\(\frac{1}{32^n}\cdot256^n=2048:2^2\)

\(=>\frac{1}{\left(2^5\right)^n}\cdot\left(2^8\right)^n=2^{10}:2^2\)

\(=>\frac{1}{2^{5n}}\cdot2^{8n}=2^8\)

\(=>2^{3n}=2^8\)

\(=>3n=8\)

\(=>n=\frac{8}{3}\)