Certainly! Let's solve the equation [tex]\( 9^x + 7 = 4 \cdot 3^x + 4 \)[/tex] step-by-step.
### Step 1: Simplify the Expression
First, observe that [tex]\( 9^x \)[/tex] can be written as [tex]\( (3^2)^x = (3^x)^2 \)[/tex]. This substitution will help simplify the equation.
So, rewrite the equation:
[tex]\[ (3^x)^2 + 7 = 4 \cdot 3^x + 4 \][/tex]
### Step 2: Introduce a Substitution
Let [tex]\( y = 3^x \)[/tex]. Substituting [tex]\( y \)[/tex] into the equation replaces [tex]\( 3^x \)[/tex] with [tex]\( y \)[/tex]. Therefore, the equation becomes:
[tex]\[ y^2 + 7 = 4y + 4 \][/tex]
### Step 3: Rearrange and Form a Quadratic Equation
Rearrange the equation to make it standard form:
[tex]\[ y^2 + 7 - 4y - 4 = 0 \][/tex]
[tex]\[ y^2 - 4y + 3 = 0 \][/tex]
### Step 4: Solve the Quadratic Equation
The quadratic equation [tex]\( y^2 - 4y + 3 = 0 \)[/tex] can be solved using the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = 3 \)[/tex].
Substituting the values, we get:
[tex]\[ y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{4 \pm \sqrt{16 - 12}}{2} \][/tex]
[tex]\[ y = \frac{4 \pm \sqrt{4}}{2} \][/tex]
[tex]\[ y = \frac{4 \pm 2}{2} \][/tex]
Thus, the solutions are:
[tex]\[ y = \frac{4 + 2}{2} = 3 \][/tex]
[tex]\[ y = \frac{4 - 2}{2} = 1 \][/tex]
### Step 5: Back-substitute [tex]\( y = 3^x \)[/tex]
Recall that [tex]\( y = 3^x \)[/tex]. Therefore, we have two equations to solve for [tex]\( x \)[/tex]:
[tex]\[ 3^x = 3 \][/tex]
[tex]\[ 3^x = 1 \][/tex]
### Step 6: Solve for [tex]\( x \)[/tex]
Solving each equation separately:
1. For [tex]\( 3^x = 3 \)[/tex]:
[tex]\[ 3^x = 3^1 \implies x = 1 \][/tex]
2. For [tex]\( 3^x = 1 \)[/tex]:
[tex]\[ 3^x = 3^0 \implies x = 0 \][/tex]
### Step 7: State the Solutions
The solutions to the equation [tex]\( 9^x + 7 = 4 \cdot 3^x + 4 \)[/tex] are:
[tex]\[ x = 0 \][/tex]
[tex]\[ x = 1 \][/tex]
So, the solutions to the equation are [tex]\( \boxed{0 \text{ and } 1} \)[/tex].
