a) \(\dfrac{32}{\left(-2\right)^n}=4\)
\(\Rightarrow\left(-2\right)^n=8=\left(-2\right)^3\)
=> n = 3
b) \(\dfrac{8}{2^n}=2\)
\(\Rightarrow2^n=4=2^2\)
=> n = 2
c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)
=> 2n - 1 = 3
=> 2n = 4
=> n = 2
Giải:
a) \(\dfrac{32}{\left(-2\right)^n}=4\)
\(\Rightarrow\left(-2\right)^n=32:4=8\)
\(\Rightarrow\left(-2\right)^n=8\)
Vì \(\left(-2\right)^n=2^3\) là ko thể nên n ∈ ∅
b) \(\dfrac{8}{2^n}=2\)
\(\Rightarrow2^n=8:2=4\)
\(\Rightarrow2^n=4\)
\(\Rightarrow2^n=2^2\)
\(\Rightarrow n=2\)
c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)
\(\Rightarrow2n-1=3\rightarrow n=2\)