What is the positive number t, if t^x+1=>3^x+2^x and positive x?

We'll create the function
f:[0;+infinte)->R

f(x)=t^x+1-2^x-3^x

We
notice that for x=0, we'll get:

f(0) = t^0 + 1 - 2^o - 3^0
= 1+1 - 1-1 = 0

We also notice that f(x) is positive for
any value of x, that is also positive.

We'll conclude that
x=0 is a minimum point.

We'll determine the 1st derivative
of f(x):

f'(x) = t^x*ln t - 2^x*ln 2 - 3^x*ln
3

Based on Fermat's theorem we'll
have:

f'(0)=0 if and only if ln t - ln 2 - ln 3 =
0

We'll move ln 2 and ln 3 to the right
side:

ln t = ln 2 + ln 3

ln t
= ln (2*3)

ln t = ln
6

t=6

We'll substitute t by 6
and we'll
get:

6^x+1=>3^x+2^x

6^x
- 2^x - 3^x + 1 >=0

2^x*3^x - 2^x - 3^x + 1
>=0

(3^x-1)(2^x-1)>=0 true, for any positive
x.

The inequality  t^x+1=>3^x+2^x is
true for t = 6.