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${{\log }_{2x}}({{x}^{2}}-5x+6)<1$

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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 doesn’t change. However, if base if less than 1, the inequality sign is reversed.

Complete step-by-step answer:

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 don’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

Complete step-by-step answer:

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 don’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