Question

7*9^(x^2-3x+1) + 5*6^(x^2-3x+1) - 48*4^(x^2-3x)

Answer 1

$7*9^{x^2-3x+1} + 5*6^{x^2-3x+1} - 48*4^{x^2-3x}=0\\7*3^2*3^{2(x^2-3x)}+5*6*2^{x^2-3x}*3^{x^2-3x}-48*2^{2(x^2-3x)}=0\\63*3^{2(x^2-3x)}+30*2^{x^2-3x}*3^{x^2-3x}-48*2^{2(x^2-3x)}=0|:3\\21*3^{2(x^2-3x)}+10*2^{x^2-3x}*3^{x^2-3x}-16*2^{2(x^2-3x)}=0|:2^{2(x^2-3x)}\\21*(\frac{3}{2})^{2(x^2-3x)}+10*(\frac{3}{2})^{x^2-3x}-16=0\\ (\frac{3}{2})^{x^2-3x}=a$

$21a^2+10a-16=0\\D:(\frac{10}{2})^2+21*16=361;\\a_1,_2=\frac{-5\pm19}{21}, \qquad a_1=\frac{2}{3}, \quad a_2=-\frac{8}{7};\\\\(\frac{3}{2})^{x^2-3x}=\frac{2}{3}\\(\frac{3}{2})^{x^2-3x}=(\frac{3}{2})^{-1}\\x^2-3x=-1\\x^2-3x+1=0\\D:9-4=5\\x_1,_2=\frac{3\pm\sqrt{5}}{2};\\\\(\frac{3}{2})^{x^2-3x} \neq -\frac{8}{7}.$