How do you solve 2log(2x) = 1 + loga?

##x=sqrt(10a)/2## (Assuming ##log = log_10##)
##2log_10(2x) = 1+log_10 a##
##2log_10 2x – log_10 a =1##
##2log_10 2x – 2log_10 a^(1/2) =1##
##2log_10((2x)/sqrt(a)) =1##
##log_10((2x)/sqrt(a)) =1/2##
##(2x)/sqrt(a) = 10^(1/2)##
##2x=sqrt(10a)##
##x=sqrt(10a)/2##