# How do you graph y>log_10(x+1)?

Dec 9, 2017

As shown in the graph

#### Explanation:

The first we consider is the graph of ${\log}_{10} \left(x + 1\right)$ this is just ${\log}_{10} \left(x\right)$ undergoing the transformation of being shifted 1 to the left:

$L o {g}_{10} \left(x\right)$ :
graph{log(x) [-1.25, 8.75, -2.64, 2.36]}

${\log}_{10} \left(x + 1\right)$: graph{log(x+1) [-2.168, 7.83, -2.62, 2.38]}

Now we must consider the inequality: Where the values of y are greater than ${\log}_{10} \left(x + 1\right)$, we see this must be above the function, where below the function the value of $y$ is smaller than the fucntion

$\implies$ graph{y>log(x+1) [-2.105, 7.893, -2.3, 2.7]}

Now we see that it cuts of at $x = - 1$ this is due to the fucntion being underfined for $x \le - 1$