Let S = \ x ∈ R: ( 3 + 2 ) ^ x + ( 3 - 2 ) ^ x = 10\ . Then the...

Given: ${\left(\sqrt{3} + \sqrt{2}\right)}^{x} + {\left(\sqrt{3} - \sqrt{2}\right)}^{x} = 10$

Now, let ${\left(\sqrt{3} + \sqrt{2}\right)}^{x} = t ⇒ {\left(\sqrt{3} - \sqrt{2}\right)}^{x} = \dfrac{1}{t}$

So, ${\left(\sqrt{3} + \sqrt{2}\right)}^{x} + {\left(\sqrt{3} - \sqrt{2}\right)}^{x} = 10$

$⇒ t + \dfrac{1}{t} = 10$

$⇒ {t}^{2} - 10t + 1 = 0$

$⇒ t = 5 \pm 2\sqrt{6}$

$⇒ t = {\left(\sqrt{3} - \sqrt{2}\right)}^{2}, {\left(\sqrt{3} + \sqrt{2}\right)}^{2}$

Now, taking ${\left(\sqrt{3} + \sqrt{2}\right)}^{x} = {\left(\sqrt{3} - \sqrt{2}\right)}^{2} = {\left(\sqrt{3} + \sqrt{2}\right)}^{- 2}$ we get, $x = - 2$

And equating ${\left(\sqrt{3} + \sqrt{2}\right)}^{x} = {\left(\sqrt{3} + \sqrt{2}\right)}^{2}$ we get, $x = 2$

Hence, the number of elements in $S$ is $2$