Question

4*3^2x +5*3^x *4^x -9*16^x =0

Answer 1

$4*3^{2x} +5*3^x *4^x -9*16^x =0\4*3^{2x} +5*3^x *4^x -9*4^{2x}=0|4^{2x}\4*(frac{3}{4})^{2x}+5*(frac{3}{4})^{x}-9=0\frac{3}{4}^x_{1,2}=frac{-5^+_-13}{8}\frac{3}{4}^x=1 ;frac{3}{4}^x=-frac{9}{4}\x=0$
(3/4)^x=-9/4 решений не имеет