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14 tháng 11 2017

2n-1:2=256

2n-1=512=29=>n-1=9=>n=10

5n+5n-2=650

5n-2(25+1)=650=>5n-2=25=52

=>n-2=2=>n=4

14 tháng 11 2017

a) \(2^{n-1}:2=256\)

\(2^{n-1-1}=256\)

\(2^{n-2}=2^8\)

\(\Rightarrow n-2=8\)

\(\Rightarrow n=10\)

vay \(n=10\)

20 tháng 9 2017

\(\dfrac{625}{5^n}\)=5

=>\(\dfrac{5^4}{5^n}\) =5

=>\(5^4\) :\(5^n\) = 5

=>\(5^{4-n}\) =\(5^1\)

=>4\(-\)n=1

=>n=4-1

=>n=3

20 tháng 9 2017

giúp mình với

14 tháng 11 2017

a)

\(2^{n-1}:2=256\)

\(\Rightarrow2^{n-1}:2=2^8\)

\(\Rightarrow2^{n-1}=2^9\)

\(\Rightarrow n-1=9\)

\(\Rightarrow n=10\)

b)

\(5^n+5^{n-2}=650\)

\(\Rightarrow5^n+5^n:5^2=650\)

\(\Rightarrow5^n+5^n:25=650\)

\(\Rightarrow5^n+5^n.\dfrac{1}{25}=650\)

\(\Rightarrow5^n.\left(1+\dfrac{1}{25}\right)=650\)

\(\Rightarrow5^n.\dfrac{26}{25}=650\)

\(\Rightarrow5^n=625\)

\(\Rightarrow5^n=5^4\)

\(\Rightarrow n=4\)

c)

\(2^{n-3}+2^{n+1}=136\)

\(\Rightarrow2^n.\dfrac{1}{2^3}+2^n.2=136\)

\(\Rightarrow2^n.\left(\dfrac{1}{8}+2\right)=136\)

\(\Rightarrow2^n.\dfrac{17}{8}=136\)

\(\Rightarrow2^n=64\)

\(\Rightarrow2^n=2^6\)

\(\Rightarrow n=6\)

14 tháng 11 2017

Thế mà "T lên chỉ để kiếm đề thôi", ngứa **t

29 tháng 10 2016

a)

\(\left(\frac{1}{3}\right)^n\cdot27^n=3^n\)

\(\Rightarrow\left(\frac{1}{3}\cdot27\right)^n=3^n\)

\(\Rightarrow9^n=3^n\)

\(\Rightarrow\left(3^2\right)^n=3^n\)

\(\Rightarrow3^{2n}=3^n\)

\(\Rightarrow2n=n\)

\(\Leftrightarrow n=0\)

Vậy \(n=0\)

29 tháng 10 2016

d) Ta có:

\(6^{3-n}=216\)

\(\Rightarrow6^{3-n}=6^3\)

\(\Rightarrow3-n=3\)

\(\Rightarrow n=3-3\)

\(\Rightarrow n=0\)

Vậy \(n=0\)\(\text{ }\)

12 tháng 2 2020

Đặt \(A=2.2^2+3.2^3+...+n.2^n\)

\(\Rightarrow2A=2.2^3+3.2^4+...+n.2^{n+1}\)

\(\Rightarrow A-2A=\)\(2.2^2+3.2^3+...+n.2^n\)\(-2.2^3-3.2^4-...-n.2^{n+1}\)

\(\Rightarrow-A=2.2^2+2^3+2^4+...+2^n-n.2^{n+1}\)

\(\Rightarrow-A=2^2+\left(2^2+2^3+2^4+...+2^{n+1}\right)-\left(n+1\right).2^{n+1}\)

\(\Rightarrow A=-2^2-\left(2^2+2^3+2^4+...+2^{n+1}\right)+\left(n+1\right).2^{n+1}\)

Đặt \(K=\left(2^2+2^3+2^4+...+2^{n+1}\right)\)

\(2K=\left(2^3+2^4+2^5+...+2^{n+2}\right)\)

\(2K-K=\left(2^3+2^4+2^5+...+2^{n+2}\right)\)\(-\left(2^2+2^3+2^4+...+2^{n+1}\right)\)

\(K=2^{n+2}-2^2\)

\(\Rightarrow A=-2^2-2^{n+2}+2^2+\left(n+1\right).2^{n+1}\)

\(\Rightarrow A=\left(n+1\right).2^{n+1}-2^{n+2}\)

\(\Rightarrow A=2^{n+1}\left(n+1-2\right)\)

\(\Rightarrow A=2^{n+1}\left(n-1\right)=2^{n+5}\Rightarrow2^4=n-1\Rightarrow n=17\)

Bài 2: 

1: \(5^n+5^{n+2}=650\)

\(\Leftrightarrow5^n\cdot26=650\)

\(\Leftrightarrow5^n=25\)

hay x=2

2: \(32^{-n}\cdot16^n=1024\)

\(\Leftrightarrow\dfrac{1}{32^n}\cdot16^n=1024\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^n=1024\)

hay n=-10

13: \(9\cdot27^n=3^5\)

\(\Leftrightarrow3^{3n}=3^5:3^2=3^3\)

=>3n=3

hay n=1

23 tháng 10 2016

\(\left(\frac{1}{2}\right)^n=\left(\frac{1}{8}\right)^5\)

\(\left(\frac{1}{2}\right)^n=\left(\frac{1^3}{2^3}\right)^5\)

\(\left(\frac{1}{2}\right)^n=\left[\left(\frac{1}{2}\right)^3\right]^5\)

\(\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^{15}\)

n = 15

23 tháng 10 2016

\(\left(\frac{1}{2}\right)^n=\left(\frac{1}{8}\right)^5\)

\(\Rightarrow\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^{3.5}\)

\(\Rightarrow\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^{15}\)

\(\Rightarrow n=15\)

Vậy n = 15