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