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#)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: \(32^{-n}.16^n=2048\)
\(\Leftrightarrow\frac{1}{2^{5n}}\cdot2^{4n}=2^{11}\)
\(\Leftrightarrow2^{4n}=2^{5n+11}\)
\(\Rightarrow4n=5n+11\)
\(\Rightarrow n=-11\)
Đề sai thì phải ! Học Lớp 7 mới giải xong bài này !
\(\frac{1}{9}\cdot27^n=3^n\)
\(\frac{1}{9}\cdot\left(3^3\right)^n=3^n\)
\(\frac{1}{9}\cdot3^{3n}=3^n\)
\(\frac{1}{9}=3^n\text{ : }3^{3n}\)
\(\frac{1}{9}=3^{-2n}\)
\(\frac{1}{3^2}=\frac{1}{3^{2n}}\)
\(\Rightarrow\text{ }3^{2n}=3^2\)
\(3^{2n}-3^2=0\)
\(3\left(3^{2n-1}-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3=0\text{ ( Vô lí ) }\\3^{2n-1}-3=0\end{cases}}\) \(\Rightarrow\text{ }3^{2n-1}=3\) \(\Rightarrow\text{ }2n-1=1\) \(\Rightarrow\text{ }2n=2\) \(\Rightarrow\text{ }n=1\)
Vậy \(n=1\)
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) \(\frac{1}{9}.27^n=3^n\)
\(\Rightarrow3^{-2}.3^{3n}=n\)
\(\Rightarrow3^{-2+3n}=n\)
\(\Rightarrow-2+3n=n\)
\(\Rightarrow2n=2\)
\(\Rightarrow n=1\)