If z∈C−{0,−2} is such thatlog(1/7)⁡|z−2|>log(1/7)⁡|z|then

log(1/7)⁡|z−2|>log(1/7)⁡|z|⇒|z−2|<|z|But |z−2|=|z| represents perpendicular bisector of the segment joining 0 and 2, that is, represents |z–2|=|z|  the line. As 0 doe  Re(z)=1 is not satisfy  we(1), get rep (1) resents Re(z) > 1