Question

The equation
$log_{2}( log_{1 \div 2}( |x| - 1) ) > 0$
has
(a) . no solution
(b) . infinite integral solution
(c) . two integral solution
(d) . infinite solutions​

Answer 1

Answer:

here's your answer

Step-by-step explanation:

hope it was help full

Answer 2

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

Given inequality is

$\rm :\longmapsto\: log_{2}( log_{ \frac{1}{2} }( |x| - 1) ) > 0$

Firstly, inequality to be defined,

$\boxed{ \rm{ \: x \: \ne \: 1,0 \: and \: - \: 1}} - - - (1)$

Now Consider

$\rm :\longmapsto\: log_{2}( log_{ \frac{1}{2} }( |x| - 1) ) > 0$

We know that,

$\purple{\boxed{ \bf{ log_{x}(y) > z \: then \: y > {x}^{z} \: iff \: x > 1}}}$

So, given inequality reduced to

$\rm :\longmapsto\: log_{ \frac{1}{2} }( |x| - 1) > {2}^{0}$

$\rm :\longmapsto\: log_{ \frac{1}{2} }( |x| - 1) > 1$

We know that,

$\purple{\boxed{ \bf{ log_{x}(y) > z \: then \: y < {x}^{z} \: iff \: 0 < x < 1}}}$

So, above inequality reduce to

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

$\rm :\longmapsto\: |x| < \dfrac{1}{2} + 1$

$\rm :\longmapsto\: |x| < \dfrac{1 + 2}{2}$

$\rm :\longmapsto\: |x| < \dfrac{3}{2}$

We know,

$\purple{\boxed{ \rm{ |x| < y\bf\implies \: - y < x < y}}}$

So,

$\rm :\implies\: - \dfrac{3}{2} < x < \dfrac{3}{2} - - - - (2)$

Combine equation (1) and (2), we get

$\bf\implies \:x \in \bigg( - \dfrac{3}{2}, - 1\bigg)\cup ( - 1,0)\cup (0,1) \cup \bigg(1,\dfrac{3}{2} \bigg)$

• So, option (d) is correct