Question

Solve the following inequalities.
$lo{g}_{1/2}lo{g}_3(1-x)>-1.$

Answer 1

We know that for ${\log }_{a}x>k$ and $a<1$

Then, $x<a^k$

${\log }_{1/2}{\log }_3(1-x)>-1$

For the given function to be defined-

$1-x>0$ and ${\log }_3(1-x)>0⟹1-x>1$

$⟹x<1$ and $x<0$

Hence for function to be defined, $x<0$

Now, for given inequality-

${\log }_{1/2}{\log }_3(1-x)>-1$

$⟹{\log }_3(1-x)<2$

$⟹1-x<9$

$⟹x>-8$

Hence final solution is

$-8<x<0$ that is $x\in (-8,0)$