Can i suggest
$\displaystyle 4^{1-7x}= 5^{x} $
$\displaystyle \ln 4^{1-7x}= \ln 5^{x} $
$\displaystyle (1-7x)\ln 4= x\ln 5 $
$\displaystyle \ln 4-7x\ln 4= x\ln 5 $
$\displaystyle \ln 4= x\ln 5+7x\ln 4 $
$\displaystyle \ln 4= x(\ln 5+7\ln 4) $
can you finish it?