Question

Solve the following inequality:${{\log }_{2x}}({{x}^{2}}-5x+6)<1$

Hint: We should take care about logarithm inequalities by taking consideration of the logarithm. That is, if the base is greater than 1, the inequality sign doesnâ€™t change. However, if base if less than 1, the inequality sign is reversed.

Thus, we should solve two cases concerning logarithm bases separately. (namely, 2x>1 and 2x<1)
Case 1: 2x > 1
\begin{align} & {{\log }_{2x}}({{x}^{2}}-5x+6)<1 \\ & {{\log }_{2x}}({{x}^{2}}-5x+6)<{{\log }_{2x}}(2x) \\ \end{align}
Here, we donâ€™t have to change the inequality sign since, 2x > 1,
\begin{align} & {{x}^{2}}-5x+6<2x \\ & {{x}^{2}}-7x+6<0 \\ & (x-1)(x-6)<0\text{ -- (1)} \\ \end{align}
From (1), we have 1 < x < 6 -- (A)
Also, since, 2x > 1
We have, x > $\dfrac{1}{2}\text{ -- (B)}$
Thus, we need to satisfy (A) and (B) together, since for Case 1 to be true, both these conditions should be true.
Thus, doing so, we would have,
1 < x < 6 (Since, this inequality satisfies both (A) and (B) both together).
Case 2: 2x < 1
\begin{align} & {{\log }_{2x}}({{x}^{2}}-5x+6)<1 \\ & {{\log }_{2x}}({{x}^{2}}-5x+6)<{{\log }_{2x}}(2x) \\ \end{align}
Here, we have to change the inequality sign since, 2x < 1,
\begin{align} & {{x}^{2}}-5x+6>2x \\ & {{x}^{2}}-7x+6>0 \\ & (x-1)(x-6)>0\text{ -- (2)} \\ \end{align}
From (2), we have, x < 1 and x > 6 -- (C)
Further, we also have, 2x < 1
So, x < $\dfrac{1}{2}\text{ -- (D)}$
Thus, we need to satisfy (C) and (D) together, since for Case 2 to be true, both these conditions should be true.
Thus, doing so, we would have,
x < $\dfrac{1}{2}$(Since, this inequality satisfies both (C) and (D) both together)
Combining the inequalities from Case 1 and Case 2, we get,
1 < x < 6 -- (I) or
x < $\dfrac{1}{2}$ -- (II)
Thus, (I) and (II) satisfy the above inequality in the problem
Note: A common mistake made is not considering case 2, where 2x < 1. Thus, it is important to remember that when solving logarithm inequalities, when the logarithm base is less than 1, the inequality sign changes. Another important thing to keep in mind is when solving a particular case (say 2x > 1), one should take the intersection of both the inequality and 2x > 1. (that is, both the inequalities should satisfy together).
To confirm the obtained inequalities, one can put the values obtained from the final answer into the question. Example, we got 1