Question

Solve the following inequalities.
$lo{g}_{0.5}(3x-1)>lo{g}_{0.5}(3-x).$

Answer 1

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

Then, $x<a^k$

Now-

${\log }_{0.5}(3x-1)>{\log }_{0.5}(3-x)$

For $\log$ to be defined-

$3x-1>0$ and $3-x>0$

$⟹x>\frac{1}{3}$ and $x<3$

Again from given inequality , we have-

$3x-1<3-x$

$⟹x<1$

Hence, overall solution is -

$\frac{1}{3}<x<1$ that is $x\in (\frac{1}{3},1)$