How do you solve for x in 2(logx-log6)=logx-2logsqrt( x - 1)?

1 Answer
Feb 3, 2016

x=(1+sqrt(145))/2 ~~6.52
(at least that's what I got)

Explanation:

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