Solve for `x` to three decimal places. `ln(x-1)^2 = 4`

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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

 

 

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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

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