2^x+7^x=5^x+4^x

[random solstice]

i saw the solution of my book:

2^x+7^x=5^x+4^x
7^x-5^x=4^x-2^x
(7^x-5^x)/(7-5)=(4^x-2^x)/(4-2) (A)
lets assume f(t)=t^x ,t>0

the equation becomes: (f(7)-f(5))/(7-5)=(f(4)-f(2))/(4-2)

so there exists ae(2,4): f'(a)=xa^(x-1)
and be(5,7): f(b)=xb^(x-1)

(A): xa^(x-1)=xb^(x-1)
x(a^(x-1)-b^(x-1))=0
x=0 or a^(x-1)=b^(x-1)

(a/b)^(x-1)=1
(a/b)^(x-1)=(a/b)^0
x-1=0
x=1