# How do you solve 2 Log (x - 4) = 4 log 2?

It is $x = 8$
2 Log (x - 4) = 4 log 2=>log(x-4)^2=log2^4=> (x-4)^2=2^4=>(x-4)=+-4=>x=8 or x=0
but hence $x > 4$ the acceptable solution is $x = 8$