Answer

Verified

456k+ views

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

Recently Updated Pages

Find the circumference of the circle having radius class 8 maths CBSE

Is the equation 6x2 + 3y 0an example of direct var class 8 maths CBSE

Why are there 2 pi radians in a circle class 8 maths CBSE

What is the remainder of 329 divided by 4 class 8 maths CBSE

The price of a pound of peppers is 399dollar What is class 8 maths CBSE

What is the greatest common divisor GCD of 78 and class 8 maths CBSE

Trending doubts

Write a letter to the principal requesting him to grant class 10 english CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

One cusec is equal to how many liters class 8 maths CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Give 10 examples of Material nouns Abstract nouns Common class 10 english CBSE

Write an application to the principal requesting five class 10 english CBSE

List out three methods of soil conservation

What is the past participle of wear Is it worn or class 10 english CBSE