Question

How do you solve `ln sqrt(x + 2) = 1`?

Answer 1

`x=e^2-2`

Explanation:

`lnsqrt(x+2)=1=>` If: `log_a(x)=yhArrx=a^y`, then:

`sqrt(x+2)=e^1 = e=>`square both sides:

`x+2=e^2=>`subtract 2 from both sides:

`x=e^2-2=>`

All roots must be checked in the original equation to verify that they work and are not "Extraneous" roots that were introduced during the squaring process, so:

`lnsqrt(x+2)=1=>` at: `x=e^2-2`
`=lnsqrt(e^2 - 2 + 2)`
`=lnsqrt(e^2)`
`=ln(e)`
`=1`
`1=1=>`hence verified: `x=e^2-2`, is a valid root.