Question

8*7^(2*x^2-x)-7*8^(2*x^2-x)=0

Answer 1

$8*7^{2x^2-x}-7*8^{2x^2-x}=0\8*7^{2x^2-x}=7*8^{2x^2-x}\(frac{7}{8} )^{2x^2-x}=frac{7}{8}\2x^2-x=1\ 2x^2-x-1=0\D=1^2+4*2=9\sqrt{D} =3\x_1=frac{1-3}{2*2}=-frac{1}{2} \x_2=frac{1+3}{2*2}=1$