Given
#color(white)("XXX")color(red)(2(log(x)-log(6))) = color(blue)(log(x)-2log(sqrt(x-1)))#
Thinks that are helpful to know:
[1]#color(white)("XXX")k*log(a) = log(a^k)#
[2]#color(white)("XXX")log(a)-log(b)=log(a/b)#
[3]#color(white)("XXX")log(sqrt(a))=log(a)/2# (this actually follows from [1])
[4]#color(white)("XXX")#for #log(a)# to be meaningful #a>0#.
#color(red)(2(log(x)-log(6)))#
#color(white)("XXX")=color(red)(2(log(x/6))# from [2]
#color(white)("XXX")=color(red)(log((x^2)/(6^2))# from [1]
#color(blue)(log(x)-2log(sqrt(x-1)))#
#color(white)("XXX")=color(blue)(log(x)-2(log(x-1)/2))# from [3]
#color(white)("XXX")=color(blue)(log(x)-log(x-1))# simplification
#color(white)("XXX")=color(blue)(log(x/(x-1))# from [2]
Therefore
#color(white)("XXX")color(red)(log(x^2/(6^2))) = color(blue)(log(x/(x-1)))#
#rArr##color(white)("XXX")x^2/36 = x/(x-1)#
#rArr##color(white)("XXX")x^3-x^2=36x#
#rArr##color(white)("XXX")x(x^2-x-36)=0#
#rArr##color(white)("XXX")x=0color(white)("XX")orcolor(white)("XX")(x^2-x-36)=0#
But #x!=0# from [4]
So
#color(white)("XXX")(x^2-x-36)=0#
This can be factored using the quadratic formula to get
#color(white)("XXX")x=(1+sqrt(145))/2color(white)("XX")orcolor(white)("XX")x=(1-sqrt(145))/2#
But #(1-sqrt(145))/2 < 0# so #x!=(1-sqrt(145))/2# from [4]
Therefore
#color(white)("XXX")x=(1+sqrt(145)/2)~~6.52#
