Question

12*4^x-35*6^x+18*9^x=0

Answer 1

$12*(2^x)^2-35*3^x*2^x+18*9^x=0 \\ \\ D=(35*3^x)^2-14*12*9^x=1225*9^x-864*9^x=361*9^x \\ \\ D=(19*3^x)^2 \\ \\ 1)2^x= \frac{35*3^x+19*3^x}{24} = \frac{54*3^x}{24} = \frac{9*3^x}{4} \\ \\ ( \frac{2}{3} )^x=( \frac{2}{3} )^{-2} \\ \\ x=-2$

$2)2^x= \frac{35*3^x-19*3^x}{24} = \frac{16*3^x}{24} = \frac{2*3^x}{3} \\ \\ ( \frac{2}{3} )^x=( \frac{2}{3} ) \\ \\ x=1 \\ \\ OTVET: {-2;1}$