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$\eqalign{ & 3^{1 - x} + 3^x \cdot 3^2 = 6\sqrt 3 \cr & \frac{3} {{3^x }} + 9 \cdot 3^x = 6\sqrt 3 \cr & 3 + 9 \cdot 3^{2x} = 6\sqrt 3 \cdot 3^x \cr & 9 \cdot 3^{2x} - 6\sqrt 3 \cdot 3^x + 3 = 0 \cr & y = 3^x \cr & 9y^2 - 6\sqrt 3 \cdot y + 3 = 0 \cr & 3y^2 - 2\sqrt 3 \cdot y + 1 = 0 \cr & \left( {\sqrt 3 \cdot y - 1} \right)^2 = 0 \cr & \sqrt 3 \cdot y = 1 \cr & y = \frac{1} {{\sqrt 3 }} = \frac{1} {3}\sqrt 3 \cr & 3^x = 3^{ - \frac{1} {2}} \cr & x = - \frac{1} {2} \cr}$