Question to study the log function.

$\log_{(f(x)-1)}{(g(x))} \leq 0 \implies 0 < f(x)-1<1$ and $g(x)\geq1$ or $f(x)-1>1$ and $0<g(x) \leq 1$

$0 < f(x)-1<1 \implies 1<\log_{|1-x|}{(x+5)}<2 \implies \log_{|1-x|}{\left(\dfrac{x+5}{|1-x|}\right)}>0 \text{ and } \log_{|1-x|}{\left(\dfrac{x+5}{(1-x)^2}\right)}<0$

$\log_{|1-x|}{\left(\dfrac{x+5}{|1-x|}\right)}>0 :$

$0<|1-x|<1$ and $0<\dfrac{x+5}{|1-x|}<1 \implies x \in (0,2) \cap(-5,-2) = \phi$ or

$|1-x|>1$ and $\dfrac{x+5}{|1-x|}>1 \implies x \in (-\infty,0)\cup(2,\infty) \cap(-2,\infty) = (-2,0) \cup (2,\infty)$

$\log_{|1-x|}{\left(\dfrac{x+5}{(1-x)^2}\right)}<0:$

$0<|1-x|<1$ and $\dfrac{x+5}{(1-x)^2}>1 \implies x \in (0,2) \cap(-1,4) = (0,2)$ or

$|1-x|>1$ and $0<\dfrac{x+5}{(1-x)^2}<1 \implies x \in (-\infty,0)\cup(2,\infty) \cap(-5,-1)\cup(4,\infty)= (-\infty,-1) \cup (4,\infty)$

$0 < f(x)-1<1 \implies x \in (-2,-1) \cup (4,\infty)$

$g(x) \geq 1 \implies x^2 - 5|x| +6 \geq 0 \implies (-\infty,-3] \cup [-2,2]\cup[3, \infty)$

$\implies x \in (-2 ,-1) \cup (4,\infty)$

$f(x)- 1>1 \implies \log_{|1-x|}{\left(\dfrac{x+5}{(1-x)^2}\right)}>0 \implies x \in (-1,0) \cup (2,4)$

$0<g(x) \leq 1 \implies x^2 - 5|x|+ 7 \leq 1 \text{ (as g(x) is always above x-axis) } \implies x \in [-3,-2] \cup[2,3]$

$\implies x \in (2,3]$

$\implies x\in (-2 ,-1) \cup (2,3] \cup (4, \infty)$

Sum of distinct solutions in $S = -1.5+2.5+3+4.5+5+5.5+6+6.5+7+7.5+8+8.5+9+9.5 = 81$

LINKS TO BRILLIANT WIKIS:

Fractional Part: {x}

Logarithmic Inequalities