\(\log _{10}\left(2\right)\left(7x+1\right)=\log _{10}\left(2\right)\left(2-x\right)\)
Expand:
\(7x\log _{10}\left(2\right)+\log _{10}\left(2\right)=2\log _{10}\left(2\right)-x\log _{10}\left(2\right)\)
\(7x\log _{10}\left(2\right)=-x\log _{10}\left(2\right)+\log _{10}\left(2\right)\)
\(8x\log _{10}\left(2\right)=\log _{10}\left(2\right)\)
\(\frac{8x\log _{10}\left(2\right)}{8\log _{10}\left(2\right)}=\frac{\log _{10}\left(2\right)}{8\log _{10}\left(2\right)}\)
So:
\(x=\frac{1}{8}\)
