Solution assignment 02 Logarithmic equations

Solve:

$\displaystyle\frac{\ln(x+1)}{4}=-1$

The domain of the function in the left-hand side is $x>-1$ .
We can rewrite the equation as:

$\ln(x+1)=-4$

$\ln(x+1)=\ln(e^{-4})$

$x+1=e^{-4}$

$x=-1+e^{-4}$

This solution is greater than $-1$ , lies in the domain and thus it is valid.