How do you solve #8x -1 = 3e^(ln(x^2))#?

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Answer 1 · 2016-06-14T12:59:58

#x=4+-sqrt15#

Let us have #a=lnb# i.e. #b=e^a# or #b=e^(lnb)#

Hence #e^(lnx^2)=x^2#

and #8x-1=e^(lnx^2)hArr8x-1=x^2# or

#x^2-8x+1=0#

Hence, using quadratic formula

#x=(-(-8)+-sqrt((-8)^2-4*1*1))/2#

or #x=(8+-sqrt(64-4))/2=(8+-2sqrt15)/2=4+-sqrt15#