Answer :
Let's solve each equation step by step.
- Solve the equation: [tex]3^{2x+1} - 26(3^x) = 9[/tex].
First, start by rewriting the equation in a way that makes it simpler to handle. Notice that [tex]3^{2x+1}[/tex] can be broken down using the properties of exponents:
[tex]3^{2x+1} = 3 \cdot 3^{2x} = 3 \cdot (3^x)^2.[/tex]
This allows you to introduce a substitution for simplicity. Let [tex]y = 3^x[/tex]. Then the equation becomes:
[tex]3y^2 - 26y = 9.[/tex]
Subtract 9 from both sides to set the equation to zero:
[tex]3y^2 - 26y - 9 = 0.[/tex]
Next, solve the quadratic equation for [tex]y[/tex]. You can use the quadratic formula:
[tex]y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},[/tex]
where [tex]a = 3[/tex], [tex]b = -26[/tex], and [tex]c = -9[/tex].
First calculate the discriminant:
[tex]b^2 - 4ac = (-26)^2 - 4 \cdot 3 \cdot (-9) = 676 + 108 = 784.[/tex]
Now substitute into the quadratic formula:
[tex]y = \frac{26 \pm \sqrt{784}}{6}.[/tex]
Since [tex]\sqrt{784} = 28[/tex], we proceed:
[tex]y = \frac{26 \pm 28}{6}.[/tex]
This gives two solutions:
- [tex]y = \frac{54}{6} = 9,[/tex]
- [tex]y = \frac{-2}{6} = -\frac{1}{3}.[/tex]
Since [tex]y = 3^x[/tex] and it's always positive, discard [tex]y = -\frac{1}{3}[/tex]. Then:
[tex]3^x = 9 \Rightarrow x = 2.[/tex]
Solution: [tex]x = 2[/tex].
- Solve the equation: [tex]9x^4 - 45x^2 = 324[/tex].
First, recognize a substitution might simplify this equation, similar to the previous one. Let [tex]z = x^2[/tex]. Then the equation transforms into:
[tex]9z^2 - 45z = 324.[/tex]
Bring all terms to one side to set the quadratic equation to zero:
[tex]9z^2 - 45z - 324 = 0.[/tex]
Now divide every term by 9 to simplify:
[tex]z^2 - 5z - 36 = 0.[/tex]
Apply the quadratic formula where [tex]a = 1[/tex], [tex]b = -5[/tex], and [tex]c = -36[/tex]:
[tex]z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.[/tex]
Calculate the discriminant:
[tex]b^2 - 4ac = (-5)^2 - 4 \cdot 1 \cdot (-36) = 25 + 144 = 169.[/tex]
Substitute into the quadratic formula:
[tex]z = \frac{5 \pm \sqrt{169}}{2}.[/tex]
Since [tex]\sqrt{169} = 13[/tex], we find:
[tex]z = \frac{5 \pm 13}{2}.[/tex]
This results in:
- [tex]z = \frac{18}{2} = 9,[/tex]
- [tex]z = \frac{-8}{2} = -4.[/tex]
Replace [tex]z = x^2[/tex] and solve for [tex]x[/tex]:
For [tex]z = 9[/tex]: [tex]x^2 = 9 \Rightarrow x = 3[/tex] or [tex]x = -3.[/tex]
For [tex]z = -4[/tex]: No real solution since squares are always non-negative.
Solutions: [tex]x = 3[/tex] and [tex]x = -3[/tex].
