# Given the function f(x)=4^x + 5^x - 3^x - 6^x show that the equation f(a)=0 has the solutions 0 and 1.

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We'll apply Lagrange's theorem to prove that the equation f(a)=0, has the solutions 0 and 1.

We'll write the equation f(a) = 0.

4^a + 5^a - 3^a - 6^a = 0

We'll keep to the left 4^a - 3^a and we'll move 5^a to the right:

4^a - 3^a = 6^a - 5^a

We'll create the function g(x) = x^a.

We'll apply Lagrange's theorem to this function, over the range [5 ; 6].

g(6)- g(5) = g'(c)(6-5)

6^a - 5^a = a*c^(a-1)

We'll apply Lagrange's theorem to this function, over the range [3 ; 4].

g(4)- g(3) = g'(d)(4-3)

4^a - 3^a = a*d^(a-1)

But 4^a - 3^a = 6^a - 5^a => a*c^(a-1) = a*d^(a-1)

We'll divide by a:

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

(c/d)^(a-1) = 1

We'll create matching bases:

(c/d)^(a-1) = (c/d)^0

This equality proves either a-1 = 0, or c = d.

The last statement, c = d, is impossible, since c is in the range (5 ; 6) and d is in the range (3 ; 4).

Only the identity a-1 = 0 holds.

a = 1

**The equation f(a) = 0 could have only 2 solutions for a: {0 ; 1}.**

f(x) = 4x+5^x-3^x-6^x.

f(a) = 4^a+5^a-3^a-6^a.

f(0) = 4^0+5^0-3^0-6^0 = 1+1-1-1 = 0.

f(1) =4^1+5^1-3^1-6^1 = 4+5-3-6 = 0.

Therefore f(a) = 0 for a= 0 and a = 1 .

So f(a) = 0 has solutions **a = 0 and a= 1.**