\(\left(\frac{1}{2}\right)^{2n-1}=\frac{1}{8}=\left(\frac{1}{2}\right)^3\)
=> 2n - 1 = 3
=> 2n = 4
=> n = 2
Vậy,........
\(\left(\frac{1}{2}\right)^{2n-1}=\frac{1}{8}\)
\(\left(\frac{1}{2}\right)^{2n-1}=\left(\frac{1}{2}\right)^3\)
\(\Rightarrow2n-1=3\)
\(\Rightarrow2n=4\Rightarrow n=2\)
\(\left(\frac{1}{2}\right)^{2n-1}=\frac{1}{8}=\left(\frac{1}{2}\right)^3\)
\(\Rightarrow2n-1=3\)
\(\Rightarrow2n=4\)
\(\Rightarrow n=2\)
Vậy n = 2.
Ta có \(\left(\frac{1}{2}\right)^{2n-1}=\frac{1}{8}\)
\(\Rightarrow\left(\frac{1}{2}\right)^{2n-1}=\left(\frac{1}{2}\right)^3\)
\(\Rightarrow2n-1=3\)
\(\Rightarrow2n=4\)
\(\Rightarrow n=2\)
( 1/2 )^2n-1 = 1/8
<=> ( 1/2 )^2n-1 = ( 1/2 )^3
<=> 2n - 1 = 3
<=> 2n = 3 + 1
<=> 2n = 4
<=> n = 4 : 2
<=> n = 2
Vậy n = 2