# Demonstrate that: 3^sqrt 5 < 5^sqrt 3 . I suppose it's quite easy and I remember I've done a lot of these comparisons, but I can't figure it out here. And I'm a little bit ashamed, since I'm in...

Demonstrate that: **3^sqrt 5 < 5^sqrt 3 . **

I suppose it's quite easy and I remember I've done a lot of these comparisons, but I can't figure it out here. And I'm a little bit ashamed, since I'm in the twelth grade :">

Thank you!

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### 3 Answers

Thank you very much! :)

To prove that 3^sqrt5 < 5^sqrt3.

Solution:

Here both sides are positive.

So we raise both sides to the same power sothat the inequality remains.

We raise both sides to the power of positive sqrt 3.

Then LHS = (3^sqrt5)^(sqrt3 = 3^(sqrt5*sqrt3) , as (a^m)^n a^mn.

So ,

(3^sqrt5 )^sqrt3 = 3^ sqrt15 < 3^sqrt16 = 3^4 = 81....(1)

RHS : (5^sqrt3)^sqrt3 = 5^(sqrt3*sqrt3) = 5^3 = 125....(2).

Therefore from (1) and (2), we get:

(3^sqrt5)^sqrt3 < 3^4 = 81 < (5^sqrt3)^sqrt3.

That proves that taking (1/sqrt) th root on both sides we get the inequality,

3^sqrt5 < 5^sqrt3 verified.

We'll apply the 4th consequence of the Lagrange's theorem.

In fact, we'll have to prove that:

n^sqrt(n+2) < (n+2)^sqrt n

We'll take natural logarithms both sides:

ln n^sqrt(n+2) < ln (n+2)^sqrt n

We've taken natural logarithms both sides, to have a base of logarithms (the base is e = 2.7...) that is > 1 and the direction of the inequality to hold.

We'll use the power property of logarithms:

sqrt(n+2)*ln n < sqrt n*ln (n+2)

We'll divide both sides by sqrt n:

sqrt(n+2)*ln n/sqrt n < ln (n+2)

We'll divide both sides by sqrt (n+2):

ln n/sqrt n < ln (n+2)/sqrt (n+2)

We'll take the function f(x) = ln x/sqrt x, where x is in the interval [3 ; 5].

If we demonstrate that the function is increasing over the interval [3 ; 5], the inequality is true.

A function is increasing if and only if, for x = 3<x = 5, we'll get:

f(3) < f(5)

Let's prove that the function is increasing. For this to hapen, the derivative of the function has to be positive. We'll do the derivative test:

f'(x) = (ln x/sqrt x)'

We'll apply the quotient rule:

(u/v)' = (u'*v-u*v')/v^2

u = ln x, u' = 1/x

v = sqrt x , v' = 1/2sqrt x

(ln x/sqrt x)' = (sqrtx/x - ln x/2sqrtx)/x

(ln x/sqrt x)' = (2sqrtx - sqrtx*lnx)/2x^2

It is obvious that (ln x/sqrt x)' > 0 over the interval [3 ; 5], so f(x) is increasing.

ln 3/sqrt 3 < ln 5/sqrt 5