Question

Tìm số nguyên dương n biết

a)\(\frac{1}{8}\).\(8^n\)=\(2^9\)

b)4.\(2^5\)≤\(2^n\)< \(4^5\)

Answer 1

a/ \(8^n\)= 4096

\(\Rightarrow\)n=4

b/ 128\(\le\)\(2^n\)<1024

\(2^7\)\(\le\)\(2^n\)<\(2^{10}\)

\(\Rightarrow\)n= \(\left\{{}\begin{matrix}7\\8\\9\end{matrix}\right.\)

Answer 2

a) \(\frac{1}{8}.8^n=2^9\)

⇔ \(\frac{8^n}{8}=2^9\)

⇔ \(\frac{2^{3n}}{2^3}=2^9\)

⇔ \(2^{3n}-2^3=2^9\)

⇔ \(3n-3=9\)

⇔ \(3n=9+3\)

⇔ \(3n=12\)

⇔ \(n=12:3\)

⇒ \(n=4\)

Vậy \(n=4.\)

b) \(4.2^5\le2^n< 4^5\)

⇔ \(2^2.2^5\le2^n< \left(2^2\right)^5\)

⇔ \(2^7\le2^n< 2^{10}\)

⇒ \(\left\{{}\begin{matrix}n=7\\n=8\\n=9\end{matrix}\right.\)

Vậy \(n\in\left\{7;8;9\right\}.\)

Chúc bạn học tốt!