To find the solution set of the inequality
3x+31−x−4<0, we begin by rewriting it as:
3x+−4<0Let
3x=t. Since
3x is always positive, we have
t>0.
Substitute into the equation:
t+−4<0Multiply the entire inequality by
t (noting that
t>0 so the inequality sign does not change):
t2−4t+3<0Factoring the quadratic, we get:
(t−3)(t−1)<0The 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)