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

1 Answer
Jul 31, 2016

Answer:

#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.