Quadratic Equations

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)