6^x-5^x = 4^x-3^x

Obviouslly when x=0

LHS:6^0-5^0 = 1-1

RHS: 4^0-3^0 = 1-1

Similarly, when x = 1,

LHS 6-5 =1

RHS 4-3 = 1.

Also for any number n, a^x - b^x = (n+a)^x - (n+b)^x holds good for x = 0 and x =1.

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To determine x, we'll use Lagrange's theorem for the function f(x)=a^x, over the intervals [3,4] and [5,6].

From Lagrange's theorem we know that there are the values c,in the interval (5,6) and d, in the interval (3,4), so that:

6^x-5^x=f'(c)(6-5)

4^x-3^x=f'(d)(4-3)

For f'(c) and f'(d), we'll calculate the first derivative having as variable c, respectively d, so, we'll differentiate power function not an exponential function.

f'(c)=xc^(x-1), f'(d)=xd^(x-1)

From enunciation, we know that:

6^x-5^x=4^x-3^x

f'(c)(6-5)=f'(d)(4-3)

f'(c)=f'(d)

xc^(x-1)= xd^(x-1)

c^(x-1)= d^(x-1)

**x=0 or x=1**