How do you solve #log_2x+log_4x=log_2 5#?

1 Answer
P dilip_k
Apr 9, 2016

#log_2x+log_4x=log_2 5#
#=>log_2x+1/log_x4=log_2 5#
#=>log_2x+1/log_x2^2=log_2 5#
#=>log_2x+1/(2log_x2)=log_2 5#
#=>log_2x+1/2xxlog_2x=log_2 5#
#=>log_2x+log_2x^(1/2)=log_2 5#
#=>log_2(x*x^(1/2))=log_2 5#
#=>(x*x^(1/2))=5#
#=>(x^(3/2))=5#
#:. x=5^(2/3)#

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