Question

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

Answer 1

`x in(1/3,oo)`

Explanation:

The function `ln(x)`, as well as any logarithmic function, has the domain `x>0`.

The same works for `ln(3x-1)`, in that we know that `3x-1>0`. Solving this yields `3x>1` and then `x>1/3`.

So, the interval for which `ln(3x-1)` is continuous is `x in(1/3,oo)`.

We can check the graph of `ln(3x-1)`:

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

The vertical asymptote approaches `x=1/3` and the function continues on towards positive infinity.