Math proof...

[jmcoreymv]

I have to show that

log (base pi) 2 + 1/ (log (base pi) 2) is greater than 2. I know the answer in words but i need math.

if you let x= (log (base pi) 2)
then you can write x+ 1/x>2

and since we know that pi^0 = 1, x must be greater than 0, and we know that pi^1=3.14..... so we know x is less then 1. And the lowest possible number you could get in that equation is 2.000000000000000(infinite)1. Since .99+1/.99>2. Any ideas how to show it mathematically.

 

[yellowplastic]

Start with x + 1/x > 2

Multiply both sides by x, hence you get x^2 + 1 > 2x.

Juggle things around, and you get x^2 - 2x + 1 > 0.

By factoring, you get (x - 1)^2 > 0.

This is true by definition, because the square of any number will be positive.

Q.E.D.

This is a trick question, you don't use log properties at all.

 

[yellowplastic]

You also have to note that since log(base pi)1 is 0, and log(base pi)pi is 1,

0 < log(base pi)2 < 1.

Then your proof is complete.

This is necessary because you would have to flip the inequality sign if you were multiplying both sides by a negative number, and you must show that x is positive in order to justify not flipping.

Also, you need to show that x is strictly a nonzero number, so that its square is also strictly nonzero.

Now we reall get Quo Erat Demonstrandum.

 

[jmcoreymv]

Thanks!

 

[b0mbrman]

log (base pi) 2 + 1/ (log (base pi) 2) > 2
(log (base pi) 2)^2 + 1 > 2 (log (base pi) 2)
(log (base pi) 2)^2 - 2 (log (base pi) 2) + 1 > 0
(log (base pi) 2) - 1)(log (base pi) 2) - 1) > 0
(log (base pi) 2) - 1)^2 > 0