Let S = {x in mathbb{R} : (sqrt{3} + sqrt{2}) | vedclass

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(C) Given equation: $(\sqrt{3} + \sqrt{2})^x + (\sqrt{3} - \sqrt{2})^x = 10$Note that $(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1$. Thus,$

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$2$

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(C) Given equation: $(\sqrt{3} + \sqrt{2})^x + (\sqrt{3} - \sqrt{2})^x = 10$Note that $(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1$. Thus,$(\sqrt{3} - \sqrt{2}) = \frac{1}{\sqrt{3} + \sqrt{2}}$.Let $t = (\sqrt{3} + \sqrt{2})^x$. Then the equation becomes $t + \frac{1}{t} = 10$.Multiplying by $t$,we get $t^2 - 10t + 1 = 0$.Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we have $t = \frac{10 \pm \sqrt{100 - 4}}{2} = \frac{10 \pm \sqrt{96}}{2} = 5 \pm 2\sqrt{6}$.Since $5 + 2\sqrt{6} = (\sqrt{3} + \sqrt{2})^2$ and $5 - 2\sqrt{6} = (\sqrt{3} - \sqrt{2})^2 = (\sqrt{3} + \sqrt{2})^{-2}$,we have $(\sqrt{3} + \sqrt{2})^x = (\sqrt{3} + \sqrt{2})^2$ or $(\sqrt{3} + \sqrt{2})^x = (\sqrt{3} + \sqrt{2})^{-2}$.Thus,$x = 2$ or $x = -2$.Therefore,the set $S = \{2, -2\}$,and the number of elements in $S$ is $2$.

Let $S = \{x \in \mathbb{R} : (\sqrt{3} + \sqrt{2})^x + (\sqrt{3} - \sqrt{2})^x = 10\}$. Then the number of elements in $S$ is:

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