The inequality in question holds for all $x\in(0,1)$.

Indeed, using the substitution $x=y^{10}$ for $y\in(0,1)$, rewrite the inequality in question as $$h(y):=\frac{1}{y^5}\,\ln\left(1-y^{10}\right)-\ln(1-y)\ge0.$$ Let $$h_1(y):=y^6 h'(y) =y^6 \left(\frac{1}{1-y}-\frac{10 y^4}{1-y^{10}}\right) -5 \ln\left(1-y^{10}\right).$$ Then $$h'_1(y)\frac{\left(1-y^{10}\right)^2}{(1-y) y^5} =5 y^{18}+9 y^{17}+12 y^{16}+14 y^{15}+15 y^{14}+65 y^{13}+64 y^{12}+62 y^{11}+59 y^{10}+55 y^9+40 y^8+26 y^7+13 y^6+y^5-10 y^4+30 y^3+21 y^2+13 y+6,$$ which is manifestly $>0$ (for $y\in(0,1)$). So, $h'_1>0$ and hence $h_1$ is increasing. Also, $h_1(0+)=0$. So, $h_1>0$ and hence $h$ is increasing. Also, $h(0+)=0$. So, $h>0$ (on $(0,1)$), as desired. $\quad\Box$

The inequality in question holds for all $x\in(0,1)$.

Indeed, using the substitution $x=y^{10}$ for $y\in(0,1)$, rewrite the inequality in question as $$h(y):=\frac{1}{y^5}\,\ln\left(1-y^{10}\right)-\ln(1-y)\ge0.$$ Let $$h_1(y):=y^6 h'(y) =y^6 \left(\frac{1}{1-y}-\frac{10 y^4}{1-y^{10}}\right) -5 \ln\left(1-y^{10}\right).$$ Then $$h'_1(y)\frac{\left(1-y^{10}\right)^2}{(1-y) y^5} =5 y^{18}+9 y^{17}+12 y^{16}+14 y^{15}+15 y^{14}+65 y^{13}+64 y^{12}+62 y^{11}+59 y^{10}+55 y^9+40 y^8+26 y^7+13 y^6 +y^3((y-5)^2+5) +21 y^2+13 y+6,$$ which is manifestly $>0$ (for $y\in(0,1)$). So, $h'_1>0$ and hence $h_1$ is increasing. Also, $h_1(0+)=0$. So, $h_1>0$ and hence $h$ is increasing. Also, $h(0+)=0$. So, $h>0$ (on $(0,1)$), as desired. $\quad\Box$