To find the solution set of the inequality 3x+31−x−4<0, we begin by rewriting it as: 3x+
3
3x
−4<0 Let 3x=t. Since 3x is always positive, we have t>0. Substitute into the equation: t+
3
t
−4<0 Multiply the entire inequality by t (noting that t>0 so the inequality sign does not change): t2−4t+3<0 Factoring the quadratic, we get: (t−3)(t−1)<0 The solution to this inequality is: t∈(1,3) Since t=3x, substitute back: 3x∈(1,3) Taking the logarithm with base 3 of both sides gives: x∈(log31,log33) Since log31=0 and log33=1, we find: x∈(0,1)