Question

Tìm n \(\varepsilon\)N :\(\frac{1}{3}.2^{n-1}+2^n=\frac{7}{3}.64\)

Answer 1

\(\frac{1}{3}.2^{n-1}+2^n=\frac{7}{3}.64\)

\(\frac{1}{3}.2^n:2^1+2^n=\frac{7}{3}.64\)

\(2^n.\frac{1}{3}.\frac{1}{2}+2^n=\frac{7}{3}.64\)

\(2^n.\frac{1}{6}+2^n.1=\frac{7}{3}.64\)

\(2^n.\left(\frac{1}{6}+1\right)=\frac{7}{3}.64\)

\(2^n.\left(\frac{1}{6}+\frac{6}{6}\right)=\frac{7}{3}.64\)

\(2^n.\frac{7}{6}=\frac{7}{3}.64\)

\(2^n=\frac{7}{3}.64:\frac{7}{6}\)

\(2^n=\frac{7}{3}.\frac{6}{7}.64\)

\(2^n=2.64\)

\(2^n=128\)

\(2^n=2^7\Rightarrow n=7\)