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a)\(\frac{1}{3^2}\cdot3^{3n}=3^n\Rightarrow3=3^{3n-2}=3^n\Rightarrow3n-2=n\Rightarrow n=1\)
b)\(\frac{1}{3^2}\cdot3^4\cdot3^n=3^7\Rightarrow3^{n+2}=3^7\Rightarrow n+2=7\Rightarrow n=5\)
a: \(5^3\cdot25^n=5^{3n}\)
\(\Leftrightarrow5^{3n}=5^3\cdot5^{2n}\)
=>3n=2n+3
hay n=3
b: \(a^{\left(2n+6\right)\left(3n-9\right)}=1\)
=>(2n+6)(3n-9)=0
=>n=-3 hoặc n=3
c: \(\dfrac{1}{3}\cdot3^n=7\cdot3^2\cdot3^4-2\cdot3^n\)
\(\Leftrightarrow3^n\cdot\dfrac{1}{3}+3^n\cdot2=7\cdot3^6\)
\(\Leftrightarrow3^n=3^7\)
hay n=7
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)\(3^{-2}.3^4.3^n=3^7\)
\(\Leftrightarrow3^{2+n}=3^7\)
\(\Leftrightarrow2+n=7\)
\(\Leftrightarrow n=5\)
câu a,
ta có: \(n\in N\)
\(32< 2^n< 128\Leftrightarrow2^5< 2^n< 2^7\)
=>n=6
câu b,
ta có:\(n\in N\)
\(2.16\ge2^n>4\\ \Leftrightarrow2.2^4\ge2^n>2^2\\ \Leftrightarrow2^5\ge2^n>2^2\\ \Rightarrow n\in\left\{5;4;3\right\}\)
câu c,
ta có:\(n\in N\)
\(\text{9 ⋅ 27 ≤ 3^n ≤ 243 }\)
\(\Leftrightarrow3^2.3^3\le3^n< 3^5\\ \Leftrightarrow3^5\le3^n< 3^5\\ \Rightarrow n\in\varnothing\)
a) \(\dfrac{1}{9}.27^n=3^n\)
\(\Leftrightarrow\dfrac{1}{9}=3^n:27^n\)
\(\Leftrightarrow\dfrac{1}{9}=\left(\dfrac{3}{27}\right)^n\)
\(\Leftrightarrow\dfrac{1}{9}=\left(\dfrac{1}{9}\right)^n\)
\(\Leftrightarrow n=1\)
b) \(3^{-2}.3^4.3^n=3^7\)
\(\Leftrightarrow3^2.3^n=3^7\)
\(\Leftrightarrow3^n=3^7:3^2\)
\(\Leftrightarrow3^n=3^5\)
\(\Leftrightarrow n=5\)
c) \(32^{-n}.16^n=2048\)
\(\Leftrightarrow\left(2^5\right)^{-n}.\left(2^4\right)^n=2^{11}\)
\(\Leftrightarrow2^{-5n}.2^{4n}=2^{11}\)
\(\Leftrightarrow2^{-n}=2^{11}\)
\(\Leftrightarrow n=-11\)