Question

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

Answer 1

$25^x-20^x-2cdot16^x<0\ (5^2)^x-(2^2cdot5)^x-2cdot(2^4)^x textless 0\ (5^x)^2-(2^2)^xcdot5^x-2cdot(2^x)^4 textless 0\ (5^x)^2-(2^x)^2cdot5^x-2cdot(2^x)^4 textless 0\ (5^x)^2+(2^x)^2cdot5^x-2cdot(2^x)^2cdot5^x-2cdot(2^x)^4 textless 0\ 5^x(5^x+(2^x)^2)-2cdot(2^x)^2(5^x+(2^x)^2) textless 0\ (5^x-2cdot(2^x)^2)(5^x+(2^x)^2) textless 0\ (5^x-2^{2x+1})(5^x+2^{2x}) textless 0\\ 5^x-2^{2x+1}=0\ 5^x=2^{2x+1}\ ln 5^x=ln (2^{2x+1})\ xln 5=(2x+1)ln 2\ xln 5=2xln 2+ln 2\ xln 5 -2xln2=ln 2\ x(ln 5-2ln 2)=ln2$
$x=dfrac{ln 2}{ln 5-2ln 2}=-dfrac{ln 2}{2ln 2-ln 5}\\\ 5^x+2^{2x}=0\ 5^x=2^{2x}\ 5^x=4^x\ xin emptyset\\ boxed{x textless -dfrac{ln 2}{2ln 2-ln 5}}$