Solve for x exponential equation: 5^x-6^x=3^x-4^x

We'll solve the equation using Lagrange's theorem on the
following intervals, [3,4] and [5,6] for the function
f(x):

f(x)=a^x

Before using
the theorem, we'll re-write the equality in this
manner:

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

According
to Lagrange's theorem, there is a value c, in the interval (5,6) and a value d, in
the interval (3,4), so
that:

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

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

Note
that we'll calculate the first derivative having as variable c, respectively d, so,
we'll differentiate a power function and not an exponential
function.

f'(c)=xc^(x-1)
(1)

f'(d)=xd^(x-1)
(2)

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

We'll
substitute both sides of equality by (1) and
(2)

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

We'll
remove the
brackets:

f'(c)=f'(d)

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

We'll divide by x both
sides:

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

x=0

or

 x=1