Logarithmic inequation

$\dfrac{\log {\dfrac{10x+2}{25(1-x)}}}{\log x}>0$

where the logarithm is in an arbitrary base.

Multiplying by $(\log x)^2$ which is always positive, $\log {\dfrac{10x+2}{25(1-x)}} \log x > 0$

Note that since $x$ is the base of the original logarithm, we have $x>0, x \neq 1$ .

Case 1: $x>1$

$\log {\dfrac{10x+2}{25(1-x)}}>0$ $\dfrac{10x+2}{25(1-x)} > 1$

Multiplying by $25(1-x)^2$ which is always positive and bringing the RHS over to the LHS, $(1-x)(35x-23)>0$ $\dfrac {23}{35}<x<1$

This case has no solution since we originally assumed $x>1$ .

Case 2: $x<1$

$\log {\dfrac{10x+2}{25(1-x)}}<0$ $\dfrac{10x+2}{25(1-x)}<1$

Multiplying by $25(1-x)^2$ which is always positive and bringing the RHS over to the LHS, $(1-x)(35x-23)<0$

Hence $x<\dfrac {23}{35}$ or $x>1$ (rejected since we originally assumed $x<1$ ).

Hence we have $0<x<\dfrac {23}{35}$ which follows from our first observation above and case 2.

Thus $x \in (0,\dfrac {161}{35*7})$ $\Rightarrow a+b = \boxed{7}$

(Note: I think the solution set should be $(0,\dfrac {23}{35})$ rather than $[0,\dfrac {23}{35}]$ .)

The better solution would be to first find the domain of x that would come out to be (0,1).

Then use this inequality ;

(x-1) {(10x+2)/(25-25x) - 1} >0

Could anyone guess how this inequality came.? 😀😀