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)))
rArrcolor(white)("XXX")x^2/36 = x/(x-1)
rArrcolor(white)("XXX")x^3-x^2=36x
rArrcolor(white)("XXX")x(x^2-x-36)=0
rArrcolor(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