How do you solve #log_6 (2x-1) + log_6 x = 2#?

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Answer 1 · 2016-06-24T06:30:22

Solution is #x=9/2#

#log_6(2x-1)+log_6x=2#

#hArrlog_6x(2x-1)=2log_6(6)=log_6(36)#

Hence #x(2x-1)=36# or #2x^2-x-36=0#

Now splitting the middle term

#2x^2-9x+8x-36=0# or

#x(2x-9)+4(2x-9)=0# or

#(x+4)(2x-9)=0#

i.e. either #x+4=0# i.e. #x=-4# (but this is out of domain)

or #2x-9=0# i.e. #x=9/2#

Hence, solution is #x=9/2#