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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: \(32^{-n}.16^n=2048\)
\(\Leftrightarrow\frac{1}{2^{5n}}\cdot2^{4n}=2^{11}\)
\(\Leftrightarrow2^{4n}=2^{5n+11}\)
\(\Rightarrow4n=5n+11\)
\(\Rightarrow n=-11\)
1.Ta có: \(\frac{x}{3}=-\frac{12}{9}\)
=> \(\frac{3x}{9}=-\frac{12}{9}\)
=> 3x = -12
=> x = -12 : 3
=> x = -4
\(\frac{4}{5}x-\frac{8}{5}=-\frac{1}{2}\)
=> \(\frac{4}{5}x=-\frac{1}{2}+\frac{8}{5}\)
=> \(\frac{4}{5}x=\frac{11}{10}\)
=> \(x=\frac{11}{10}:\frac{4}{5}\)
=> \(x=\frac{11}{8}\)
\(\frac{1}{2}.2^n+4.2^n=9.2^5\)
\(2^n.\left(\frac{1}{2}.4\right)=288\)
\(2^n.2=288\)
\(2^n=288:2\)
\(2^n=144\)
Suy ra n ko tìm được
Ta có :
\(\frac{1}{2}\cdot2^n+4\cdot2^n=\frac{9}{2}\cdot2^5\)
\(=>\left(\frac{1}{2}+4\right)\cdot2^n=9\cdot2^5\)
\(=>\left(\frac{1}{2}+\frac{8}{2}\right)\cdot2^n=9\cdot2^5\)
\(=>\frac{9}{2}\cdot2^n=\frac{9}{2}\cdot2^5\)
\(=>2^n=2^5\)
\(=>n=5\)