Question

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

Answer 1

`x=e-1`

Explanation:

In general `ln(a)=c` means `e^c=a`
therefore
`color(white)("XXX")ln(x+1)=1`
means
`color(white)("XXX")e^1=x+1`

`color(white)("XXX")rarr x=e-1`