Let's solve the equation [tex]\( e^{2x} - 8e^x + 12 = 0 \)[/tex].
First, we notice that the equation has an exponential component, which can make it challenging to solve directly. To simplify this, let's make a substitution. Define [tex]\( u = e^x \)[/tex]. This transforms the original equation into a quadratic equation in terms of [tex]\( u \)[/tex]:
[tex]\[
u^2 - 8u + 12 = 0
\][/tex]
Next, we solve this quadratic equation using the quadratic formula, which is given by:
[tex]\[
u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
In our equation [tex]\( u^2 - 8u + 12 = 0 \)[/tex], we have:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = -8 \)[/tex]
- [tex]\( c = 12 \)[/tex]
Plugging these values into the quadratic formula, we get:
[tex]\[
u = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 12}}{2 \cdot 1}
\][/tex]
Simplify the equation inside the square root:
[tex]\[
u = \frac{8 \pm \sqrt{64 - 48}}{2}
\][/tex]
[tex]\[
u = \frac{8 \pm \sqrt{16}}{2}
\][/tex]
[tex]\[
u = \frac{8 \pm 4}{2}
\][/tex]
This gives us two possible solutions for [tex]\( u \)[/tex]:
[tex]\[
u = \frac{8 + 4}{2} = \frac{12}{2} = 6
\][/tex]
[tex]\[
u = \frac{8 - 4}{2} = \frac{4}{2} = 2
\][/tex]
Therefore, the solutions to the equation [tex]\( u^2 - 8u + 12 = 0 \)[/tex] are [tex]\( u = 6 \)[/tex] and [tex]\( u = 2 \)[/tex].
Since [tex]\( u = e^x \)[/tex], we now revert back to the original variable [tex]\( x \)[/tex]:
1. [tex]\( e^x = 6 \)[/tex]
2. [tex]\( e^x = 2 \)[/tex]
Solving for [tex]\( x \)[/tex] in each case, we take the natural logarithm (ln) of both sides:
1. [tex]\( x = \ln(6) \)[/tex]
2. [tex]\( x = \ln(2) \)[/tex]
Evaluating these expressions, we approximate the solutions as:
1. [tex]\( x \approx 1.79175946922805 \)[/tex]
2. [tex]\( x \approx 0.693147180559945 \)[/tex]
Thus, the solutions to the equation [tex]\( e^{2x} - 8e^x + 12 = 0 \)[/tex] are:
[tex]\[
x \approx 1.79175946922805 \quad \text{and} \quad x \approx 0.693147180559945
\][/tex]
