Find the solutions of the exponential equation
e^2x–4e^x+3=0
x1=? x2=?
let u = e^x and you have
(u-1)(u-3) = 0
so e^x = 1 or e^x = 3
got it?
what happens to the 2 in e^2x?
do u bring it down?
(e^x)^2 = e^2x
To find the solutions of the given exponential equation, e^2x - 4e^x + 3 = 0, we can use a substitution to simplify the equation.
Let's make a substitution: Let u = e^x. This means that e^2x = u^2.
Now, we can rewrite the equation using the substitution:
u^2 - 4u + 3 = 0
This is now a quadratic equation in terms of u. We can solve it by factoring or using the quadratic formula.
Factoring:
(u - 1)(u - 3) = 0
Setting each factor equal to zero and solving for u gives us:
u - 1 = 0 --> u = 1
u - 3 = 0 --> u = 3
Since u = e^x, we can substitute u back in to get the values of x:
e^x = 1 --> x = ln(1) = 0 (natural logarithm of 1 is 0)
e^x = 3 --> x = ln(3)
Therefore, the solutions to the exponential equation e^2x - 4e^x + 3 = 0 are:
x1 = 0
x2 = ln(3)