Question

Показательное уравнение.

Answer 1

$(2+\sqrt{3})^{7-x}=(\frac{1}{7+4\sqrt{3}})^{2x+1}\\\\(2+\sqrt{3} )^{7-x}=(\frac{1}{2^{2}+4\sqrt{3}+(\sqrt{3})^{2}})^{2x+1}\\\\(2+\sqrt{3} )^{7-x}=(\frac{1}{(2+\sqrt{3})^{2}})^{2x+1}\\\\(2+\sqrt{3})^{7-x}=((2+\sqrt{3})^{-2})^{2x+1}\\\\(2+\sqrt{3})^{7-x}=(2+\sqrt{3})^{-4x-2}\\\\7-x=-4x-2\\\\-x+4x=-2-7\\\\3x=-9\\\\x=-3\\\\Otvet:\boxed{-3}$