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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 doesnâ€™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 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

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 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

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