# How do you find the intervals on which the function is continuous given  y = ln(3x-1)?

Jul 31, 2016

$x \in \left(\frac{1}{3} , \infty\right)$

#### Explanation:

The function $\ln \left(x\right)$, as well as any logarithmic function, has the domain $x > 0$.

The same works for $\ln \left(3 x - 1\right)$, in that we know that $3 x - 1 > 0$. Solving this yields $3 x > 1$ and then $x > \frac{1}{3}$.

So, the interval for which $\ln \left(3 x - 1\right)$ is continuous is $x \in \left(\frac{1}{3} , \infty\right)$.

We can check the graph of $\ln \left(3 x - 1\right)$:

graph{ln(3x-1) [-15.21, 20.84, -9.05, 8.97]}

The vertical asymptote approaches $x = \frac{1}{3}$ and the function continues on towards positive infinity.