Question

Simplify the given equation ${\log }_{5}x+{\log }_x\frac{x}{3}<\frac{{\log }_{5}x(2-{\log }_{3}x)}{{\log }_{3}x}$

Answer 1

${\log }_{5}x+{\log }_x\left(\frac{x}{3}\right)<{\log }_{5}x\frac{(2-{\log }_{3}x)}{{\log }_{3}x}$

$\frac{\log x}{\log 5}+\frac{\log x}{\log x}-\frac{\log 3}{\log x}<2\frac{\log x}{\log 5}\frac{\log 3}{\log x}-{\log }_{5}x$

$\frac{\left(\log x\right)}{\left(\log 5\right)}+1-\frac{\log 3}{\left(\log x\right)}<2\frac{\log 3}{\log 5}-\frac{\log x}{\log 5}$

$2\frac{\log x}{\log 5}-\frac{\log 3}{\log x}<2\frac{\log 3}{\log 5}-1$

$2\frac{{\left(\log x\right)}^2-\left(\log 3\right)\left(\log 5\right)}{\left(\log x\right)\left(\log 5\right)}<2\frac{\log 3-\log 5}{\left(\log 5\right)}$

$\left(\log x\right)[2{\left(\log x\right)}^2-\left(\log 3\right)\left(\log 5\right)]<(2\log 3-\log 5){\left(\log x\right)}^2$

$[2{\left(\log x\right)}^2-(2\log 3-\log 5)\log x-\left(\log 3\right)\left(\log 5\right)]\left(\log c\right)<0$

$[2\log x+1\log 5][\log x-\log 3]\left(\log x\right)<0$

$\log xϵ(-\infty ,-\frac{1}{2}\log 5)\bigcup (\log 1,\log 3)$

$xϵ(0,\frac{1}{25})\bigcup (1,3)$