Помогитеее пожалуйста

Решите задачу:

$(49\sqrt{3} )^0 = 1 = (\frac{7}{3} )^0\\x^2+2x-15 = 0\\\frac{D}{4} = 1 + 15 = 16\\x1 = -1 +4 = 3\\x2 = -1-4 = -5\\(3-5)/2 = -1$

$2*2^x * 81^x - 9*36^x = 0\\2*162^x-9*36^x=0\\(\frac{9}{2})^x = \frac{9}{2} \\x = 1$

$2*7^{x+1} +5 = 7^{-x}\\2*7^{2x+1}+5*7^{x}=1\\14*7^{2x}+5*7^x-1=-\\D = 25+56=81\\7^{x} = (-5 +9)/28=1/7\\7^{x}=(-5-9)/28=-1/2\\\\x = -1$

$4*3^{2x}+5*3^x*4^x-9*16^x=0\\4*9^x+5*12^x-9*16^x=0\\4*(\frac{9}{16}) ^{x} + 5*(\frac{3}{4})^{x} - 9 = 0\\ D = 25+144=169\\(\frac{3}{4} )^x = (-5 +13)/8 = 1; (\frac{3}{4}) ^x = (-5-13)/8 = -\frac{9}{4} \\x = 0$