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

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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.**