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a) \(2^{-1}\cdot2^n+4\cdot2^n=9\cdot2^5\)
\(\Rightarrow2^n\cdot\left(2^{-1}+4\right)=9\cdot2^5\)
\(\Rightarrow2^n\cdot4,5=288\)
\(\Rightarrow2^n=64\)
\(\Rightarrow n=6\)
b) \(2^m-2^n=1984\)
\(\Rightarrow2^n\cdot\left(2^{m-n}-1\right)=2^6\cdot31\)
\(\Rightarrow\left\{{}\begin{matrix}2^n=2^6\\2^{m-n}-1=31\end{matrix}\right.\)
\(\Rightarrow n=6\)
\(\Rightarrow2^{m-n}=32\Rightarrow m-n=5\Rightarrow m=11\)
a) \(\left(\frac{1}{3}\right)^n=\frac{1}{81}\)
\(\Rightarrow\left(\frac{1}{3}\right)^n=\frac{1^4}{3^4}\)
\(\Rightarrow\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow n=4\)
Vậy n = 4
b) \(\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow\frac{-8^3}{7^3}=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow\left(\frac{-8}{7}\right)^3=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow n=3\)
Vậy n = 3
\(a. \)
\(\dfrac{-32}{\left(-2\right)^n}=4\)
\(\Rightarrow\left(-2\right)^n=-32:4=-8\)
\(\Rightarrow\left(-2\right)^n=\left(-2\right)^3\)
\(\Rightarrow n=3\)
\(b.\)
\(\dfrac{8}{2^n}=2\)
\(\Rightarrow2^n=4\)
\(\Rightarrow2^n=2^2\)
\(\Rightarrow n=2\)
\(c.\)
\(\dfrac{16}{\left(-2\right)^n}=-8\)
\(\Rightarrow\left(-2\right)^n=-2\)
\(\Rightarrow n=1\)
a) \(2^m+2^n=2^{m+n}\)
\(\Leftrightarrow2^m+2^n=2^m.2^n\)
\(\Leftrightarrow2^m.2^n-2^m-2^n=0\)
\(\Leftrightarrow2^m\left(2^n-1\right)-\left(2^n-1\right)=1\)
\(\Leftrightarrow\left(2^m-1\right)\left(2^n-1\right)=1=1.1=\left(-1\right).\left(-1\right)\)
\(TH1:\hept{\begin{cases}2^m-1=1\\2^n-1=1\end{cases}}\Leftrightarrow\hept{\begin{cases}m=1\\n=1\end{cases}}\)
\(TH1:\hept{\begin{cases}2^m-1=-1\\2^n-1=-1\end{cases}}\Leftrightarrow m,n\in\left\{\varnothing\right\}\)
Vậy m = n = 1
\(2^m-2^n=256\)
\(\Leftrightarrow2^n\left(2^{m-n}-1\right)=2^8\)
\(TH1:m-n< 2\)\(\Rightarrow\hept{\begin{cases}n=8\\m=9\end{cases}}\)
\(TH2:m-n\ge2\)
VP chứa toàn thừa số nguyên tố 2 nên VP chẵn.
*Xét VT: \(2^{m-n}-1\)lẻ vì \(m-n\ge2\)
Suy ra : VT lẻ, VP chẵn ( vô lí )
Vậy m = 9 , n = 8