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.

Question

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.

William Abdalla · Accepted Answer

See below.

We must make a comparison. #(sqrt(a))^(sqrt(b)-1)# and #(sqrt(b))^(sqrt(a)-1)# Applying #log# to both sides #(sqrt b - 1)log(sqrt(a)) > (sqrta -1)log(sqrt(b))# or #log(sqrt(a))/(sqrta - 1) > log(sqrtb)/(sqrtb - 1)# The function #g(x) = log(x)/(x-1)# is strictly monotonic decreasing so if #a > b# then #g(b) > g(a)#. Here #a = sqrt(2) < b = sqrt(3)# then #g(2) > g(3)# as proposed.

Samantha Adkison · Answer

undefined 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:

$\sqrt{2}^{\sqrt{3}-1} > \sqrt{3}^{\sqrt{2}-1}$

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

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.