`ln(x^2 - 2) = ln(23)` Use the One-to-One Property to solve the equation for `x`.

Textbook Question

Chapter 3, 3.2 - Problem 75 - Precalculus (3rd Edition, Ron Larson).
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loves2learn | (Level 3) Salutatorian

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Since there are ln's on both sides, set the terms equal to each other.

`x^2-2=23`

Solve for x

`x^2-25=0 `

Factor

`(x+5)(x-5)=0 `

`x=-5 ` and `x=5 `

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