\(\frac{1}{9}.3^4.3^n=3^7\)
Tìm n
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Mk làm lun, ko viết lại đề bài nữa nhé =))
a) \(\Leftrightarrow\)\(3^2.3^{n+1}=9^4\)
\(\Leftrightarrow3^{n+1}=9^4:3^2\)
\(\Leftrightarrow3^{n+1}=3^6\)
\(\Rightarrow n+1=6\)
\(\Leftrightarrow n=6-1\)
\(\Rightarrow n=5\)
b)\(\Leftrightarrow2^n.\left(\frac{1}{2}+4\right)=9.2^5\)
\(\Leftrightarrow2^n.\frac{9}{2}=9.2^5\)
\(\Rightarrow2^n=\left(9.2^5\right):\frac{9}{2}\)
\(\Rightarrow2^n=468:\frac{9}{2}\)
Tự tính nốt KQ giúp mk nha ♥
a) 32 . 3n = 35
=> 3n = 35 : 32
=> 3n = 33
=> n = 3
b) (22 : 4) . 2n = 4
=> (4 : 4) . 2n = 4
=> 2n = 4
=> 2n = 22
=> n = 2
c) \(\frac{1}{9}.3^4.3^n=3^7\)
\(\Rightarrow3^{-2}.3^4.3^n=3^7\)
\(\Rightarrow3^{-2+4+n}=3^7\)
\(\Rightarrow3^{2+n}=3^7\)
\(\Rightarrow2+n=7\)
\(\Rightarrow n=5\)
d) \(\frac{1}{9}.27^n=3^n\)
\(\Rightarrow3^{-2}.3^{3n}=n\)
\(\Rightarrow3^{-2+3n}=n\)
\(\Rightarrow-2+3n=n\)
\(\Rightarrow2n=2\)
\(\Rightarrow n=1\)
#)Giải :
\(\frac{1}{9}.3^4.3^n=3^7\)
\(\frac{1}{9}.81.3^n=3^7\)
\(9.3^n=3^7\)
\(3^2.3^n=3^7\)
\(\Rightarrow2+n=7\)
\(\Rightarrow n=5\)
#~Will~be~Pens~#
a) Ta có: \(\frac{1}{9}\cdot27^n=3^n\)
\(\Leftrightarrow\frac{1}{3^2}\cdot\left(3^3\right)^n=3^n\)
\(\Leftrightarrow3^{3n}=3^{n+2}\)
\(\Rightarrow3n=n+2\)
\(\Rightarrow n=1\)
b) Ta có: \(3^2.3^4.3^n=3^7\)
\(\Rightarrow3^n=3\)
\(\Rightarrow n=1\)
c) Ta có: \(2^{-1}.2^n+4.2^n=9.2^5\)
\(\Leftrightarrow2^n\cdot\frac{9}{2}=9.2^5\)
\(\Rightarrow2^n=2^6\)
\(\Rightarrow n=6\)
d) Ta có: \(32^{-n}.16^n=2048\)
\(\Leftrightarrow\frac{1}{2^{5n}}\cdot2^{4n}=2^{11}\)
\(\Leftrightarrow2^{4n}=2^{5n+11}\)
\(\Rightarrow4n=5n+11\)
\(\Rightarrow n=-11\)
a) \(\frac{1}{9}.27^n=3^n\)
\(\Leftrightarrow3^{-2}.3^{3n}=3^n\)
\(\Leftrightarrow3^{3n-2}=3^n\)
\(\Leftrightarrow3n-2=n\)
\(\Leftrightarrow2n=2\)
\(\Leftrightarrow n=1\)
b)\(3^{-2}.3^4.3^n=3^7\)
\(\Leftrightarrow3^{2+n}=3^7\)
\(\Leftrightarrow2+n=7\)
\(\Leftrightarrow n=5\)
\(\frac{1}{9}\cdot3^4\cdot3^n=3^8\)
\(=>3^n=3^8:3^4:\frac{1}{9}\)
\(=>3^n=3^8:3^4\cdot9\)
\(=>3^n=3^8:3^4\cdot3^2\)
\(=>3^n=3^6\)
\(=>n=6\)
b) \(\frac{1}{9}.3^4.3^n=3^8\)
\(\Rightarrow\left(\frac{1}{3}\right)^2.3^4.3^n=3^8\)
\(\Rightarrow\frac{1}{3^2}.3^4.3^n=3^8\)
\(\Rightarrow3^2.3^n=3^8\)
\(\Rightarrow3^n=3^8:3^2\)
\(\Rightarrow3^n=3^6\)
\(\Rightarrow n=6\)
Vậy n = 6
\(\frac{1}{9}.3^4.3^n=3^7\)
=) \(\frac{1}{3^2}.3^4.3^n=3^7\)=) \(3^2.3^n=3^7\)
=) \(3^{2+n}=3^7\)=) \(2+n=7\)
=) \(n=7-2\)=) \(n=5\)
Ta có: \(\frac{1}{9}.3^4.3^n=3^{-2}.3^4.3^n=3^{-2+4+n}=3^7\)
\(\Rightarrow-2+4+n=7\)
\(\Rightarrow2+n=7\)
\(\Rightarrow n=5\)