Question

send the solution fast​

Answer 1

Step-by-step explanation:

We have,

$\log_{2}( {x}^{2} - 2x) < 0$

$\implies{x}^{2} - 2x < {2}^{0}$

$\implies{x}^{2} - 2x < 1$

$\implies{x}^{2} - 2x - 1 < 0$

$\implies \: \{x - ( 1 - \sqrt{2} ) \}\{x - (1 + \sqrt{2} ) \} < 0$

$\implies \: ( 1 - \sqrt{2} ) < x < (1 + \sqrt{2})$

Also,

according to the defination of log,

${x}^{2} - 2x > 0$

$\implies \: x(x - 2) > 0$

$\implies \: x \in( - \infty, 0) \: \cup \: (2, \infty )$

So,

$\tt \red{x \in(1 - \sqrt{2},0) \: \cup \: (2 ,1 + \sqrt{2} ) }$

Answer 2

$\large\underline{\sf{Given \:Question - }}$

The solution set of Logarithmic inequality is

$\rm :\longmapsto\: log_{2}( {x}^{2} - 2x) < 0$

$\large\underline{\sf{Solution-}}$

The given Logarithmic inequality is

$\rm :\longmapsto\: log_{2}( {x}^{2} - 2x) < 0$

Let first define the domain of the above function.

We know,

$\boxed{ \rm \:logx \: is \: defined \: when \: x > 0}$

So,

$\rm :\longmapsto\: {x}^{2} - 2x > 0$

$\rm :\longmapsto\:x(x - 2) > 0$

$\bf\implies \:x < 0 \: \: or \: \: x > 2$

$\bf\implies \:x \: \in \: ( - \infty , \: 0) \: \cup \: (2, \: \infty ) - - (1)$

Now,

$\rm :\longmapsto\: log_{2}( {x}^{2} - 2x) < 0$

$\rm :\longmapsto\: {x}^{2} - 2x < {2}^{0}$

$\rm :\longmapsto\: {x}^{2} - 2x < 1$

$\rm :\longmapsto\: {x}^{2} - 2x - 1 < 0$

$\rm :\longmapsto\: {x}^{2} - (1 + 1)x - 1 < 0$

$\rm :\longmapsto\: {x}^{2} - (1 + 1 + \sqrt{2} - \sqrt{2} )x - (2 - 1) < 0$

$\rm :\longmapsto\: {x}^{2} - (\sqrt{2} + 1 + 1- \sqrt{2} )x - ( {( \sqrt{2}) }^{2} - 1) < 0$

$\rm :\longmapsto\: {x}^{2} - ( \sqrt{2} + 1)x + ( \sqrt{2} - 1)x - ( \sqrt{2} + 1)( \sqrt{2} - 1) < 0$

$\rm :\longmapsto\:x(x - \sqrt{2} - 1) + ( \sqrt{2} - 1)(x - \sqrt{2} - 1) < 0$

$\rm :\implies\:(x - 1 - \sqrt{2})(x + \sqrt{2} - 1) < 0$

$\rm :\implies\:1 - \sqrt{2} < x < 1 + \sqrt{2}$

$\bf\implies \:x \: \in \: \bigg(1 - \sqrt{2}, \: 1 + \sqrt{2} \bigg) - - - (2)$

From equation (1) and (2), we concluded that

$\bf\implies \:x \: \in \: \bigg(1 - \sqrt{2}, \: 0\bigg) \: \cup \: \bigg(2, \: 1 + \sqrt{2} \bigg)$

Result used :-

If a < b, then

$\boxed{ \rm \:(x - a)(x - b) < 0 \: \: \bf\implies \:a < x < b}$

$\boxed{ \rm \:(x - a)(x - b) > 0 \: \: \bf\implies \:x < a \: \: or \: \: x > b}$

$\boxed{ \rm \: log_{x}(y) < z, \: and \: x > 1 \: then \: y < {x}^{z}}$