Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\frac{-32}{\left(-2\right)^n}=4\)
\(\frac{\left(-2\right)^5}{\left(-2\right)^n}=4\)
\(\left(-2\right)^{5-n}=\left(-2\right)^2\)
=> 5-n = 2
n = 3
b) \(\frac{8}{2^n}=2\)
\(\frac{2^3}{2^n}=2\)
\(2^{3-n}=2^1\)
=> 3 -n = 1
n = 2
c) \(\left(\frac{1}{2}\right)^{2n-1}=\frac{1}{8}\)
\(\left(\frac{1}{2}\right)^{2n-1}=\left(\frac{1}{2}\right)^3\)
=> 2n -1 = 3
2n = 4
n = 2
a) \(\frac{-32}{\left(-2\right)^n}=4\Leftrightarrow\left(-2\right)^n=\frac{-32}{4}\)
\(\left(-2\right)^n=-8\)Mà \(-8=2^{-3}\)
\(\Rightarrow x=-3\)
b) \(\frac{8}{2^n}=2\Leftrightarrow2^n=\frac{8}{2}\)
\(2^n=4\) Mà \(4=2^2\Rightarrow x=2\)
c) \(\left(\frac{1}{2}\right)^{2n-1}=\frac{1}{8}\Rightarrow\left(\frac{1}{2}\right)^{2n}:\frac{1}{2}=\frac{1}{8}\)
\(\left(\frac{1}{2}\right)^{2n}=\frac{1}{8}\cdot\frac{1}{2}\)
\(\left(\frac{1}{2}\right)^{2n}=\frac{1}{16}\Leftrightarrow\frac{1}{2^{2n}}=\frac{1}{16}\) mà\(16=2^4\)
\(2n=4\Rightarrow n=2\)
Vậy .........................
c, \(\frac{-32}{-2^n}=4\)
\(\Rightarrow-2^n=-32:4\)
\(\Rightarrow-2^n=-8\)
\(\Rightarrow-2^n=-2^3\Rightarrow n=3\)
d, \(\frac{8}{2^n}=2\)
\(\Rightarrow2^n=8:2\)
\(\Rightarrow2^n=4\)
\(\Rightarrow2^n=2^2\Rightarrow n=2\)
e, \(\frac{25^3}{5^n}=25\)
\(\Rightarrow5^n=25^3:25\)
\(\Rightarrow5^n=25^2\)
\(\Rightarrow5^n=5^4\Rightarrow n=4\)
i , \(8^{10}:2^n=4^5\)
\(\Rightarrow2^n=8^{10}:4^5\)
\(\Rightarrow2^n=\left(2^3\right)^{10}:\left(2^2\right)^5\)
\(\Rightarrow2^n=2^{30}:2^{10}\)
\(\Rightarrow2^n=2^{20}\Rightarrow n=20\)
k, \(2^n.81^4=27^{10}\)
\(\Rightarrow2^n=27^{10}:81^4\)
\(\Rightarrow2^n=\left(3^3\right)^{10}:\left(3^4\right)^4\)
\(\Rightarrow2^n=3^{30}:3^{16}\)
\(\Rightarrow2^n=3^{14}\)
\(\Rightarrow2^n=4782969\)Không chia hết cho 2 nên ko có Gt n thỏa mãn
Bài 1: Tìm n biết
a) -32 / -2^n= 4
\(-\frac{32}{-2^n}=4\Rightarrow\frac{32}{2^n}=4\Rightarrow4\cdot2^n=32\)
\(\Rightarrow2^n=8\Rightarrow2^n=2^3\Rightarrow n=3\)
b) ( 1/2 ) ^2n-1 = 1/8
Ta có : \(\left(\frac{1}{2}\right)^{2n-1}=\frac{1}{8}\)
mà \(\frac{1}{8}=\left(\frac{1}{2}\right)^3\)
\(\Rightarrow2n-1=3\)
\(\Rightarrow2n=4\)
\(\Rightarrow n=2\)
Bài 2:Tìm x
a)
\(\frac{x}{\frac{4}{2}}=\frac{4}{\frac{x}{2}}\Rightarrow\frac{x}{4}=\frac{4}{x}\)
\(\Rightarrow x^2=4\cdot4\)
\(\Rightarrow x^2=16\)
mà 16=42=(-4)2
\(\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
Vậy \(x\in\left\{4;-4\right\}\)
b) (x+5)^3 = -64
Vì (x+5)3 = -64
mà -64=(-4)3
\(\Rightarrow x+5=-4\)
\(\Rightarrow x=1\)
c) (2x-3)^2 = 9
\(\Rightarrow\orbr{\begin{cases}2x-3=3\\2x-3=-3\end{cases}\Rightarrow\orbr{\begin{cases}2x=6\\2x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=0\end{cases}}}\)
Vậy \(x\in\left\{3;0\right\}\)
\(2n-1+5n-2=\frac{7}{32}\)
\(\Rightarrow\left(2n+5n\right)-\left(1+2\right)=\frac{7}{32}\)
\(\Rightarrow7n-3=\frac{7}{32}\)
\(\Rightarrow7n=\frac{53}{96}\)
\(\Rightarrow n=\frac{53}{672}\)
Mà \(n=\frac{53}{672}\notin Z\)
\(\Rightarrow x\) không có giá trị thỏa mãn
Vậy \(x\) không có giá trị thỏa mãn
1. \(\left(\frac{1}{2}\right)^n=\frac{1}{32}\)
\(\left(\frac{1}{2}\right)^n=\frac{1^5}{2^5}\)
\(\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^5\)
Vậy \(n=5\)
2. \(\frac{343}{125}=\left(\frac{7}{5}\right)^n\)
\(\frac{7^3}{5^3}=\left(\frac{7}{5}\right)^n\)
\(\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\)
Vậy \(n=3\)
3. \(\frac{16}{2^n}=2\)
\(2^n=\frac{16}{2}\)
\(2^n=8=2^3\)
Vậy \(n=3\)
1. (1/2)2 = 1/32 <=> (21)n = (25)n <=> 1.n = 5.1 <=> n = 5
=> n = 5
2) 343/125 = (7/5)n <=> (7/5)3 = (7/5)n <=> 3 = n
=> n = 3
3) 16/2n = 2 <=> 16.2n <=> 2n = 2/16 <=> 2n = 1/8 <=> 2n = 8 <=> 2n = 23 <=> n = 3
=> n = 3
\(-\frac{32}{\left(-2\right)^n}=4\\ =>-\frac{32}{\left(-2\right)^n}=\frac{4}{1}\\ =>\left(-2\right)^n=8\\ =>n=3\)
\(-\frac{32}{\left(-2\right)^n}=4\Rightarrow\frac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)
=> (-2)5:(-2)n=(-2)2
=> 5-n=2
=> n=3
\(\frac{-32}{-2^n}=4\)
\(\Leftrightarrow-2^n=-8\)
\(\Leftrightarrow n=3\)
\(\frac{-32}{-2^n}=4\)
\(\frac{32}{2^n}=4\)
\(\frac{2^5}{2^n}=2^2\)
\(2^{5-n}=2^2\)
5 - n = 2
n =3