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

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Answer 1 · 2016-08-15T01:56:03

#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#