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

Jun 15, 2016

#### Explanation:

As the domain for $\ln x$ is $x > 0$, the domain for $\ln \left(1 + x\right)$ is $x > - 1$. However range is $- \infty < \ln \left(1 + x\right) < \infty$.

Function is continuous and relation is one to one.

Again for $0 < x < 1$, $\ln x$ is negative and hence $\ln \left(1 + x\right)$ is negative for $- 1 < x < 0$.

And as $x \to - 1$, $\ln \left(1 + x\right) \to - \infty$ and hence $x + 1 = 0$ is asymptote for $\ln \left(1 + x\right)$.

At $x = 0$, $\ln \left(1 + x\right) = 0$ (it also means $\ln \left(1 + x\right)$ cuts $x$-axis at $\left(0.0\right)$ and for $x > 0$ $\ln \left(1 + x\right) > 0$.

One can also take additional values for $x$ such as $\left\{1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10\right\}$ for which $\ln \left(1 + x\right)$ would be $\left\{0.693 , 1.099 , 1.386 , 1.609 , 1.792 , 1.946 , 2.079 , 2.197 , 2.303 , 2.398 , 2.485\right\}$. The graph of $\ln \left(1 + x\right)$ appears as given below.

graph{ln(1+x) [-5.415, 14.585, -6, 4]}