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##

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