To solve the equation [tex]\( 3^{x^2 + 6x} = 9^{-4} \)[/tex], let's follow these detailed steps:
1. Rewrite the Right-Hand Side:
Notice that 9 can be written as [tex]\( 3^2 \)[/tex]. Thus, [tex]\( 9^{-4} \)[/tex] can be rewritten using base 3:
[tex]\[
9^{-4} = (3^2)^{-4}
\][/tex]
Applying the power of a power rule [tex]\((a^m)^n = a^{mn}\)[/tex], we get:
[tex]\[
9^{-4} = 3^{2 \cdot (-4)} = 3^{-8}
\][/tex]
2. Set Exponents Equal:
Now that both sides of the equation have the same base (3), we can equate the exponents:
[tex]\[
3^{x^2 + 6x} = 3^{-8} \quad \Longrightarrow \quad x^2 + 6x = -8
\][/tex]
3. Form a Quadratic Equation:
The above equation is a standard quadratic equation. Rewrite it in the standard form:
[tex]\[
x^2 + 6x + 8 = 0
\][/tex]
4. Solve the Quadratic Equation:
To solve [tex]\( x^2 + 6x + 8 = 0 \)[/tex], we use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = 8 \)[/tex].
First, compute the discriminant:
[tex]\[
b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot 8 = 36 - 32 = 4
\][/tex]
Next, calculate the values of [tex]\( x \)[/tex]:
[tex]\[
x = \frac{-6 \pm \sqrt{4}}{2 \cdot 1} = \frac{-6 \pm 2}{2}
\][/tex]
This gives us two solutions:
[tex]\[
x = \frac{-6 + 2}{2} = \frac{-4}{2} = -2
\][/tex]
[tex]\[
x = \frac{-6 - 2}{2} = \frac{-8}{2} = -4
\][/tex]
5. Conclusion:
The solutions to the equation [tex]\( 3^{x^2 + 6x} = 9^{-4} \)[/tex] are:
[tex]\[
x = -2, -4
\][/tex]
