We have to solve e^2x - 6*e^x + 5 = 0 for x.

Let e^x = y

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

=> y^2 - 5y - y + 5 = 0

=> y(y - 5) - 1(y - 5) = 0

=> (y - 1)(y - 5) = 0

=> y = 1 and y =5

e^x = 1 and e^x = 5

=> x = 0 and x = ln 5

**The solution of the equation are x = 0 and x = ln 5**

The equation e^2x-6*e^x+5=0 has to be solved.

e^2x-6*e^x+5=0

= (e^x)^2-6*e^x+5=0

= (e^x)^2-5*e^x - e^x +5=0

= e^x*(e^x - 5) - 1(e^x - 5)=0

= (e^x - 1)*(e^x - 5)=0

e^x - 1 = 0

e^x = 1

This is the case when x = 0

e^x - 5 = 0

e^x = 5

Take the natural logarithm of both the sides

x = ln 5

The solution of the equation e^2x-6*e^x+5=0 is x = 0 and x = ln 5

We'll solve the equation using the technique of substitution.

We'll substitute e^x by the variable t.

e^2x=(e^x)^2=t^2

We'll re-write the equation in t:

t^2 -6t + 5 = 0

We'll apply the quadratic formula:

t1 = [6+sqrt(36-20)]/2

t1 = (6+4)/2

t1 = 5

t2 = (6-4)/2

t2 = 1

Now, we'll determine x:

e^x = t1

e^x = 5

We'll apply natural logarithms boths sides:

ln e^x = ln 5

x = ln 5

e^x = t2

e^x = 1

But e^0 = 1

e^x = e^0

x = 0

**The requested solutions of the equation are: {0 ; ln 5}.**