# Solve the equation.e^2x - 6e^x + 5 = 0

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We have to solve: e^2x - 6e^x + 5 = 0

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

let e^x = t

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

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

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

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

=> e^x = 1 and e^x = 5

=> x = 0 and x = ln 5

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

We'll apply substitution technique to solve the exponential equation.

e^x = t

We'll raise to square both sides:

e^2x = t^2

t^2 - 6t + 5 = 0

We'll apply quadratic formula:

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

t1 = (6 + 4)/2

t1 = 5

t2 = 1

But e^x = t1 => e^x = 5

We'll take natural logarithms both sides:

ln e^x = ln 5

x*ln e = ln 5

x = ln 5

e^x = t2=> e^x = 1

x = 0

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