Question

Find 'x' for which $lo{g}_{1/2}x>lo{g}_{1/3}x$.

• A. $\frac{1}{3}<x<\frac{1}{2}$
• B. $x>\frac{1}{2}$
• C. 0 < x < 1
• D. $x<\frac{1}{3}$ and $x>\frac{1}{2}$

Answer 1

The correct option is C 0 < x < 1

From the inequality, we have
$lo{g}_{1/2}x-lo{g}_{1/3}x=lo{g}_{1/2}x-lo{g}_{1/2}x.lo{g}_{1/3}x.\frac{1}{2}>0$
$\Rightarrow lo{g}_{1/2}x[1-lo{g}_{3^2}]>0$
[Since $lo{g}_{1/3}1/2=lo{g}_3^{-1^{2^{-1}}}=lo{g}_{3}2]$
$\Rightarrow lo{g}_{1/2}x>0$, As $lo{g}_{3}2<1$,
$1-lo{g}_{3}2>0$
But $0<base<1$
$\therefore$ 0 < x < 1