Solve for `x` to three decimal places. `ln(x-1)^2 = 4`
3 Answers
flbyrne | (Level 3) Assistant Educator
Solve `ln((x-1)^2)=4`
Apply inverse logarithm
`(x-1)^2=e^4`
Take the first square root:
`x-1=e^2`
`x=e^2+1`
`=2.71828^2+1=8.389`
Take the second root:
` ` `-x+1=e^2`
`x=1-e^2=-6.389`
The solutions are x=8.389 and x=-6.389
ishpiro | College Teacher | (Level 1) Educator
First, rewrite the left side of the equaton using the rules of logarithms:
`ln(x-1)^2 = 2ln|x-1|`
Then,
`2ln|x-1| = 4`
Divide both sides by 2:
`ln|x-1| = 2`
In the exponential form, this states that `|x-1| = e^2`
Asbsolute value eqation breaks up into two equations:
`x-1 = e^2` and `x-1 = -e^2`
From here, `x = e^2 + 1 = 8.389`
and `x = 1 - e^2 = -6.389`
So, the solutions of the given equation, to three decimal places are
x = 8.389 and x = -6.389
`ln(x-1)^2=4`
So , by definition of ln (natural log)
`(x-1)^2=e^4`
`=>`
`x-1=+-sqrt(e^4)`
`x-1==+-e^2`
`x-1=+-7.389`
`x=8.389 or -6.389`