Let's start by analyzing and solving the given exponential equation step-by-step:
[tex]\[
3^{2x} - 4 \times 3^{x+1} + 27 = 0
\][/tex]
First, let's make a substitution to simplify the equation. Let:
[tex]\[
y = 3^x
\][/tex]
This implies that:
[tex]\[
3^{2x} = (3^x)^2 = y^2
\][/tex]
Also, we can simplify:
[tex]\[
4 \times 3^{x+1} = 4 \times 3^x \times 3 = 4 \times 3 \times y = 12 y
\][/tex]
Substituting these into the original equation, we get:
[tex]\[
y^2 - 12y + 27 = 0
\][/tex]
Now, we have a quadratic equation in terms of [tex]\( y \)[/tex]. The general form of a quadratic equation is:
[tex]\[
ay^2 + by + c = 0
\][/tex]
In this case, [tex]\( a = 1 \)[/tex], [tex]\( b = -12 \)[/tex], and [tex]\( c = 27 \)[/tex].
To solve the quadratic equation, we can use the quadratic formula:
[tex]\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Plugging in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[
y = \frac{12 \pm \sqrt{(-12)^2 - 4 \times 1 \times 27}}{2 \times 1}
\][/tex]
[tex]\[
y = \frac{12 \pm \sqrt{144 - 108}}{2}
\][/tex]
[tex]\[
y = \frac{12 \pm \sqrt{36}}{2}
\][/tex]
[tex]\[
y = \frac{12 \pm 6}{2}
\][/tex]
This gives us two solutions for [tex]\( y \)[/tex]:
[tex]\[
y = \frac{12 + 6}{2} = 9
\][/tex]
[tex]\[
y = \frac{12 - 6}{2} = 3
\][/tex]
Recalling our substitution [tex]\( y = 3^x \)[/tex], we now have:
[tex]\[
3^x = 9 \quad \text{or} \quad 3^x = 3
\][/tex]
Solving for [tex]\( x \)[/tex]:
1. For [tex]\( 3^x = 9 \)[/tex]:
[tex]\[
3^x = 3^2 \Rightarrow x = 2
\][/tex]
2. For [tex]\( 3^x = 3 \)[/tex]:
[tex]\[
3^x = 3^1 \Rightarrow x = 1
\][/tex]
Thus, the solutions for the equation [tex]\( 3^{2x} - 4 \times 3^{x+1} + 27 = 0 \)[/tex] are:
[tex]\[
x = 1 \quad \text{and} \quad x = 2
\][/tex]
