Certainly! Let's solve the given equation step-by-step:
[tex]\[
(2^{2n+1})(3^{2n+2}) + 2^n (3^{n+2}) - 2 = 0
\][/tex]
### Step 1: Simplify the terms
First, let's simplify the given terms for clarity.
1. First Term:
[tex]\[
(2^{2n+1})(3^{2n+2})
\][/tex]
This can be written as:
[tex]\[
2^{2n+1} \cdot 3^{2n+2}
\][/tex]
2. Second Term:
[tex]\[
2^n (3^{n+2})
\][/tex]
This can be written as:
[tex]\[
2^n \cdot 3^{n+2}
\][/tex]
### Step 2: Factor out common terms (if any)
Notice that there is a common factor of [tex]\(2^n \cdot 3^{n+2}\)[/tex] in both exponential terms.
So let's factor out [tex]\(2^n \cdot 3^{n+2}\)[/tex]:
[tex]\[
2^n \cdot 3^{n+2} \left( 2^{2n+1-n} \cdot 3^{2n+2-(n+2)} \right) + 2^n \cdot 3^{n+2} - 2 = 0
\][/tex]
This simplifies to:
[tex]\[
2^n \cdot 3^{n+2} \left( 2^{n+1} \cdot 3^n + 1 \right) - 2 = 0
\][/tex]
### Step 3: Isolate the term involving [tex]\(n\)[/tex] if possible
Now, observe the equation:
[tex]\[
2^n \cdot 3^{n+2} \left( 2^{n+1} \cdot 3^n + 1 \right) - 2 = 0
\][/tex]
We can rewrite this equation by setting [tex]\( k = 2^n \cdot 3^{n+2} \)[/tex]:
[tex]\[
k \left( 2^{n+1} \cdot 3^n + 1 \right) - 2 = 0
\][/tex]
### Step 4: Solve for [tex]\(n\)[/tex]
To solve for [tex]\(n\)[/tex], we consider possible values for the simplified form of the equation. Through analysis, we find the solutions:
1. First Solution: [tex]\( n = -1 \)[/tex]
This value directly satisfies the given equation when substituted back into the original terms.
2. Second Solution: [tex]\( n = \frac{\log(2/3) + i\pi}{\log(6)} \)[/tex]
This comes from analyzing the exponential and logarithmic properties of the given equation.
### Conclusion:
Therefore, the solutions to the equation [tex]\((2^{2n+1})(3^{2n+2}) + 2^n (3^{n+2}) - 2 = 0\)[/tex] are:
[tex]\[
n = -1
\][/tex]
and
[tex]\[
n = \frac{\log(2/3) + i\pi}{\log(6)}
\][/tex]
These are the two values of [tex]\(n\)[/tex] that satisfy the given equation.
