# How to demonstrate that #sqrt(2)^(sqrt(3)-1)>sqrt(3)^(sqrt(2)-1)#?We know that #g:(1,+oo)->RR,g(x)=##(lnx)/(x-1)#,and #g# is decreasing.

See below.

We must make a comparison.

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To demonstrate that (\sqrt{2}^{\sqrt{3}-1} > \sqrt{3}^{\sqrt{2}-1}), we'll use the given function (g(x) = \frac{\ln x}{x-1}), where (g: (1, +\infty) \rightarrow \mathbb{R}) and (g(x)) is decreasing.

First, rewrite the inequality using the given function:

(\frac{\ln(\sqrt{2})}{\sqrt{2}-1} > \frac{\ln(\sqrt{3})}{\sqrt{3}-1})

Next, we'll simplify:

(\frac{\ln(2)}{\sqrt{2}-1} > \frac{\ln(3)}{\sqrt{3}-1})

Now, we can substitute (x = \sqrt{2}) and (y = \sqrt{3}) into the function (g(x)):

(g(\sqrt{2}) > g(\sqrt{3}))

Since (g(x)) is decreasing, this implies:

(\sqrt{2}^{\sqrt{3}-1} > \sqrt{3}^{\sqrt{2}-1})

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