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

$-1/3 < x <3$