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yes I did that and i got
7 over 1024 = 28(1over2)to the power of n - 1

and i divided each sides by 28

and i used log method and I am not getting a right answer..

\[(n-1) \log(\frac{1}{2}) = \log(\frac{7}{1024*28})\]
\[n-1 = 12\]
n = 13

ohhhhh okay thanks!!! :)