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

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hala718 eNotes educator| Certified Educator

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

Let us substitute with c=0:

==> 6^0 - 5^0 = 4^0 - 3^0

==> 1- 1 = 1-1 = 0

Then:

x1= 0  is one of the solutions:

Now subsotute with x=1:

==> 6^1 - 5^1 = 4^1 - 3^1

==> 6-5 = 4-3

==> 1= 1

Then:

x2= 1 is asolution:

Then x= { 0, 1}

neela | Student

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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giorgiana1976 | Student

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

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