Sure, let's solve the equation step-by-step:
We have the equation:
[tex]\[ 3^{2x} - 10 \cdot 3^x + 3^2 = 0 \][/tex]
First, we will make a substitution to simplify the equation. Let:
[tex]\[ u = 3^x \][/tex]
Then the equation becomes:
[tex]\[ u^2 - 10u + 9 = 0 \][/tex]
Now, we need to solve for [tex]\( u \)[/tex] using the quadratic formula. The quadratic formula is given by:
[tex]\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In this case, [tex]\( a = 1 \)[/tex], [tex]\( b = -10 \)[/tex], and [tex]\( c = 9 \)[/tex]. Plugging in these values, we get:
[tex]\[ u = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1} \][/tex]
[tex]\[ u = \frac{10 \pm \sqrt{100 - 36}}{2} \][/tex]
[tex]\[ u = \frac{10 \pm \sqrt{64}}{2} \][/tex]
[tex]\[ u = \frac{10 \pm 8}{2} \][/tex]
This results in two solutions for [tex]\( u \)[/tex]:
[tex]\[ u = \frac{10 + 8}{2} = 9 \][/tex]
[tex]\[ u = \frac{10 - 8}{2} = 1 \][/tex]
Now, we need to revert our substitution [tex]\( u = 3^x \)[/tex]:
[tex]\[ 3^x = 9 \][/tex]
[tex]\[ 3^x = 1 \][/tex]
Let's solve each one separately:
1. For [tex]\( 3^x = 9 \)[/tex]:
Since [tex]\( 9 \)[/tex] can be written as [tex]\( 3^2 \)[/tex], we have:
[tex]\[ 3^x = 3^2 \][/tex]
Thus:
[tex]\[ x = 2 \][/tex]
2. For [tex]\( 3^x = 1 \)[/tex]:
Since [tex]\( 1 \)[/tex] can be written as [tex]\( 3^0 \)[/tex], we have:
[tex]\[ 3^x = 3^0 \][/tex]
Thus:
[tex]\[ x = 0 \][/tex]
So, the solutions to the equation [tex]\( 3^{2x} - 10 \cdot 3^x + 9 = 0 \)[/tex] are:
[tex]\[ x = 0 \quad \text{and} \quad x = 2 \][/tex]
