You should equate the exponents such that:

-6=(x^2-5x)

You should move all terms to the left side such that:

-x^2 + 5x - 6 = 0

You need to multiply by -1 such that:

x^2 - 5x + 6 = 0

You should complete the square such that:

x^2 - 5x + 25/4 - 25/4 + 6 = 0

(x - 5/2)^2 = 25/4 - 6 => (x - 5/2)^2 = 1/4

x - 5/2 = +-1/2

x - 5/2 = 1/2 => x = (5+1)/2 => x = 3

x - 5/2 = -1/2 => x = (5-1)/2 => x = 2

**Hence, evaluating the solutions to exponential equation yields x = 2 ; x = 3.**

We have to find x given that 5^-6=5^(x^2-5x)

5^-6=5^(x^2-5x)

=> 1/5^6 = 5^(x^2 - 5x)

=> 1 = 5^6*5^(x^2 - 5x)

=> 1 = 5^(6 + x^2 - 5x)

As 5^0 = 1

=> x^2 - 5x + 6 = 0

=> x^2 - 3x - 2x + 6 = 0

=> x(x - 3) - 2(x - 3) = 0

=> (x - 2)(x - 3) = 0

=> x = 2 and x = 3

**The solution of the equation is x = 2 and x = 3**

This is an exponential equation.

We'll write the equation again using symmetric property:

5^(x^2-5x) = 5^-6

We'll use the one to one property, because the bases are matching:

x^2 - 5x = -6

We'll add 6 both sides:

x^2 - 5x + 6 = 0

We'll apply the quadratic formula:

x1 = [5+sqrt(25 - 24)]/2

x1 = (5+1)/2

x1 = 3

x2 = (5-1)/2

x2 = 2

Verifying:

For x1 = 3:

5^(3^2-5*3) = 5^-6

5^(9-15) = 5^-6

5^-6 = 5^-6

For x2 = 2:

5^(2^2-5*2) = 5^-6

5^(4-10) = 5^-6

5^-6 = 5^-6

We'll not reject any of found solutions.