How do you solve #log_10(4x)-log_10(12+sqrtx)=2#?

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Answer 1 · 2017-02-05T14:32:10

#x=(125(10+sqrt73))/2#

As #loga-logb=log(a/b)#

#log_10(4x)-log_10(12+sqrtx)=2# is equivalent to

#log_10((4x)/(12+sqrtx))=2#

Hence from definition of log, we have

#((4x)/(12+sqrtx))=100#

or #4x-100sqrtx-1200=0# and dividing by #4#

#x-25sqrtx-300=0# and using quadratic formula

#sqrtx=(25+-sqrt(625+1200))/2=(25+-5sqrt73)/2#

But we cannot have #4x<0#, hence #sqrtx=(25+5sqrt73)/2#

#x=(25+5sqrt73)^2/4=(625+1825+250sqrt73)/4#

or #(2500+250sqrt73)/4#

or #(250(10+sqrt73))/4#

or #(125(10+sqrt73))/2#