Question #69781

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Answer 1 · 2017-01-24T14:12:10

The answer is #-1/3 < x <3#

#log(3x+1)<1#

By definition for #log#

#(3x+1)>0#

#3x> -1#

#x> -1/3#

Now, let's solve the inequation

#log(3x+1)<1#

#(3x+1)<10^1#

#3x<10-1#

#3x<9#

#x<3#

Putting it all together

#-1/3 < x <3#