Question

Solve the following inequality:
$\frac{1}{5-{\log }_{a}x}+\frac{2}{1+{\log }_{a}x}<1$

Answer 1

$\frac{1}{(5-{\log }_{a}x)}+\frac{2}{(1+{\log }_{a}x)}<1$

$-\frac{1}{({\log }_{a}x-5)}+\frac{2}{({\log }_{a}x+1)}<1$

$\frac{-{\log }_{a}x-1+2{\log }_{a}x-10}{({\log }_{a}x+1)({\log }_{a}x-5)}<1$

$\frac{({\log }_{a}x-11)}{({\log }_{a}x+1)({\log }_{a}x-5)}<1$

For $a>1,({\log }_{a}x+1)({\log }_{a}x-5)<0$

For $xϵ(\frac{1}{a},a^5)$

and $({\log }_{a}x+1)({\log }_{a}x-5)>0$

for $xϵ(-\infty ,\frac{1}{a})\bigcup (a^5,\infty )$

For $xϵ(-\infty ,\frac{1}{a})\bigcup (a^5,\infty ),a>1$

$({\log }_{a}x-11)<{\log }_a{}^{2}x-4{\log }_{a}x-5$

${\log }_a{}^{2}x-5{\log }_{a}x+6>0$

$({\log }_{a}x-2)({\log }_{a}x-3)>0$

${\log }_{a}xϵ(-\infty ,2)\bigcup (3,\infty )$

$xϵ(0,a^2)\bigcup (a^3,\infty )$

Hence

$xϵ(0,\frac{1}{a})\bigcup (a^5,\infty )$

For $xϵ(\frac{1}{a},a^5),a>1$

$({\log }_{a}x-1)>({\log }_{a}x+1)({\log }_{a}x-5)$

${\log }_a{}^{x}x-5{\log }_{a}x+6<0$

${\log }_{a}xϵ(2,3)$

$xϵ(a^2,a^3)$

Hence, $xϵ(a^2,a^3)$

Taking union $xϵ(0,\frac{1}{a})\bigcup (a^2,a^3)\bigcup (a^5,\infty )$

For $aϵ(0,1)(a=\frac{1}{k})$

$(-{\log }_{k}x-11)<1$

$(-{\log }_{k}x+1)(-{\log }_{k}x-5)$

$\frac{-\left({\log }_{k}x+11\right)}{({\log }_{k}x-1)({\log }_{k}x+5)}<1$

For $xϵ(\frac{1}{k^5},k)$

$-({\log }_{k}x+11)>({\log }_{k}x-1)({\log }_{k}x+5)$

$-({\log }_{k}x+11)>{\log }_k{}^{2}x+4{\log }_{k}x-5$

${\log }_k{}^{2}x+5{\log }_{k}x+6<0$

$({\log }_{k}x+2)({\log }_{k}x+3)<0$

$xϵ(\frac{1}{k^3},\frac{1}{k^2})$

Hence, $xϵ(\frac{1}{k^3},\frac{1}{k^2})$

$xϵ(a^3,a^2)$

For $xϵ(-\infty ,\frac{1}{k^5})\bigcup (k,\infty )$

$-({\log }_{k}x+11)<({\log }_{k}x-1)({\log }_{k}x+5)$

$-({\log }_{k}x+11)<{\log }_k{}^{2}x+4{\log }_{k}x-5$

${\log }_k{}^{2}x+5{\log }_{k}x+6>0$

$({\log }_{k}x+2)({\log }_{k}x+3)>0$

$xϵ(-\infty ,\frac{1}{k^3})\bigcup (\frac{1}{k^2},\infty )$

Hence $xϵ(-\infty ,\frac{1}{k^5})\bigcup (k,\infty )$

$xϵ(-\infty ,a^5)\bigcup (a,\infty )$

For $aϵ(0,1)$

$xϵ(a^3,a^2)\bigcup (-\infty ,a^5)\bigcup (a,\infty )$, where a is in fraction.