# Solve equation for x: a) `ln(x^2-1)=3`

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To solve this equation, we need to change the equation from logarithmic to exponential form.

`ln(x^2-1)=3`

`x^2-1=e^3` move 1 to other side

`x^2=1+e^3` take square roots

`x=+-sqrt{1+e^3}`

For both potential solutions, when we substitute back into the original equation, we get valid solutions.

**The solutions of the equation are `x=+-sqrt{1+e^3}` .**