# How do you graph y=ln(x+1)?

Shift the graph of $\ln \left(x\right)$ to the left by $1$
Remember that since $\ln \left(x\right)$ and ${e}^{x}$ are inverse functions $\ln \left({e}^{x}\right) = x$.
Because if $y = \ln \left(x\right)$ then $y = 0$ if and only if $\ln \left(x\right) = 0$. Since ${e}^{0} = 1$ then $\ln \left({e}^{0}\right) = \ln \left(1\right) = 0.$ So when we change the function to $y = \ln \left(x + 1\right)$, we have that $y = 0 = \ln \left(1\right) = \ln \left(0 + 1\right)$ So the x-intercept shifts to the left. Just like the x-intercept shift, the entire graph shifts to the left.