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\(6.2^n+3.2^n=9.2^9\)
\(\left(6+3\right).2^n=9.2^9\)
\(9.2^n=9.2^9\)
\(\Rightarrow2^n=2^9\)
\(\Rightarrow n=9\)
Vậy \(n=9\)
\(2^n:4=16\)
\(2^n=16.4\)
\(2^n=64\)
\(2^n=2^6\)
\(\Rightarrow n=6\)
có 6.2n+3.2n =2n.(6+3)= 9.2n
=>9.2n=9.29
2n=29 .9:9
2n=29
=>n=9
a) \(2^n:4=16\Rightarrow2^n:2^2=2^4\Rightarrow2^{n-2}=2^4\Rightarrow n-2=4\Rightarrow n=6\)
b) \(6\cdot2^n+3\cdot2^n=9\cdot2^9\)
=> \(\left(6+3\right)\cdot2^n=9\cdot2^9\)
=> \(9\cdot2^n=9\cdot2^9\Rightarrow n=9\)
c) \(3^n:3^2=243\)
=> \(3^{n-2}=3^5\)
=> n - 2 = 5 => n = 7
d) 25 < 5n < 3125
=> 52 < 5n < 55
=> n \(\in\){3;4}
\(a,3^2\cdot3^4\cdot3^n=3^{12}\)
\(\Rightarrow3^{6+n}=3^{12}\)
\(\Rightarrow6+n=12\)
\(\Rightarrow n=6\)
\(b,2^n:4=16\)
\(\Rightarrow2^n:2^2=2^4\)
\(\Rightarrow2^{n-2}=2^4\)
\(\Rightarrow n-2=4\)
\(\Rightarrow n=6\)
\(c,6\cdot2^n+3\cdot2^n=9\cdot2^9\)
\(\Rightarrow2^n\left(6+3\right)=9\cdot2^9\)
\(\Rightarrow2^n\cdot9=9\cdot2^9\)
\(\Rightarrow2^n=2^9\)
\(\Rightarrow n=9\)
a)
\(3^4.3^n:9=3^7\)
\(\Rightarrow3^4.3^n=3^7.9\)
\(\Leftrightarrow3^4.3^n=3^7.3^2\)
\(\Rightarrow3^4.3^n=3^9\)
\(\Rightarrow3^n=3^9:3^4\)
\(\Rightarrow3^n=3^5\)
\(\Rightarrow n=5\)
Vậy \(n=5\)
d)
\(2^n:4=16\)
\(\Leftrightarrow2^n:2^2=2^4\)
\(\Rightarrow2^n=2^4.2^2\)
\(\Rightarrow2^n=2^6\)
\(\Rightarrow n=6\)
Vậy \(n=6\)
a)\(9^3.3^n=3^{12}\Rightarrow\left(3^2\right)^3.3^n=3^{12}\Rightarrow3^6.3^n=3^{12}\Rightarrow3^n=3^{12}:3^6=3^2\)\(\Rightarrow n=2\)
b)\(\left(2n+4\right)^2-5.7=4^2-15\)
\(\left(2n+2^2\right)^2-35=2^4-15\)
\(2n^2+2^4=2^4-15+35\)
\(2n^2+2^4=2^4+20\)
\(2n^2=20\)
mà 20 k fai số chính phương nên k tìm đc n
c)\(\left(n-2\right)^5=243\Rightarrow\left(n-2\right)^5=3^5\Rightarrow n-2=3\Rightarrow n=5\)
d)\(\left(n+1\right)^3=125\Rightarrow\left(n-1\right)^3=5^3\Rightarrow n-1=5\Rightarrow n=6\)
e)\(6.2^n+3.2^n=9.2^2\)
\(2^n\left(3+6\right)=9.2^2\)
\(2^n.9=9.2^2\Rightarrow2^n=2^2\Rightarrow n=2\)
Mk thấy mấy bài này cx đâu có khó j đâu, bn chỉ cần vận dụng công thức là đc thôi mà
****nha
\(6\times2^n+3\times2^n=9\times2^9\)
\(2^n\times\left(6+3\right)=9\times2^9\)
\(2^n\times9=9\times2^9\)
\(n=9\)
ta có:\(2^n.\left(6+3\right)=9.2^9\)
\(2^n.9=2^9.9\)
\(\Rightarrow2^n=2^9\)
\(\Rightarrow n=9\)
Vậy n=9