#log_11(2x-1)=1-log_11(x+4)# What is #x#?

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Answer 1 · 2017-12-17T13:40:03

#log_11(2x-1)=1-log_11(x+4)#

#=>log_11(2x-1)+log_11(x+4)=1#

#=>log_11[(2x-1)(x+4)]=log_11 11#
#=>(2x-1)(x+4)]= 11#

#=>2x^2-x+8x-4- 11=0#

#=>2x^2+7x-15=0#

#=>2x^2+10x-3x-15=0#

#=>2x(x+5)-3(x+5)=0#

#=>(x+5)(2x-3)=0#

So #x=-5 and x=3/2#

But #x=-5# is not valid for the given equation #log_11(2x-1)=1-log_11(x+4)#

Hence #x=3/2#