Log[1/7](5x+3)>=-1/2

$log_{ \frac{1}{7} }(5x+3)= log_{ 7^{-1} }(5x-3)=- log_{7}(5x-3)$
$\frac{1}{2}= log_{7} 7^{ \frac{1}{2} }$
-log₇(5x+3)>=-log₇(√7) => log₇(5x+3)<=log₇(√7) => 5x+3<=√7 => x<=<img src="https://tex.z-dn.net/?f=+%5Cfrac%7B+%5Csqrt%7B7%7D-3+%7D%7B5%7D+" id="TexFormula3" title=" \frac{ \sqrt{7}-3 }{5} " alt=" \frac{ \sqrt{7}-3 }{5} " align="absmiddle" class="latex-formula">
5x+3>0 => x>-3/5
-3/5<x<=(√7-3)/5