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

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To solve for the equation :

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

We can see that x = 0 gives:

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

RHS = 3^0-4^0 =0

So the equation satisfies for x = 0.

Similarly for x =1 ,

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

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

Therefore x = 0 and x =1 are the two roots of the given exponential equation.

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**