25 в степени Х - 20 в степени х - 2*16 в степени х<0

Решите задачу:

$25^x-20^x-2\cdot16^x<0\\ (5^2)^x-(2^2\cdot5)^x-2\cdot(2^4)^x\ \textless \ 0\\ (5^x)^2-(2^2)^x\cdot5^x-2\cdot(2^x)^4\ \textless \ 0\\ (5^x)^2-(2^x)^2\cdot5^x-2\cdot(2^x)^4\ \textless \ 0\\ (5^x)^2+(2^x)^2\cdot5^x-2\cdot(2^x)^2\cdot5^x-2\cdot(2^x)^4\ \textless \ 0\\ 5^x(5^x+(2^x)^2)-2\cdot(2^x)^2(5^x+(2^x)^2)\ \textless \ 0\\ (5^x-2\cdot(2^x)^2)(5^x+(2^x)^2)\ \textless \ 0\\ (5^x-2^{2x+1})(5^x+2^{2x})\ \textless \ 0\\\\ 5^x-2^{2x+1}=0\\ 5^x=2^{2x+1}\\ \ln 5^x=\ln (2^{2x+1})\\ x\ln 5=(2x+1)\ln 2\\ x\ln 5=2x\ln 2+\ln 2\\ x\ln 5 -2x\ln2=\ln 2\\ x(\ln 5-2\ln 2)=\ln2\\$
$x=\dfrac{\ln 2}{\ln 5-2\ln 2}=-\dfrac{\ln 2}{2\ln 2-\ln 5}\\\\\\ 5^x+2^{2x}=0\\ 5^x=2^{2x}\\ 5^x=4^x\\ x\in \emptyset\\\\ \boxed{x\ \textless \ -\dfrac{\ln 2}{2\ln 2-\ln 5}}$