To solve the equation $3^{2x} + 9 = 10 \cdot 3^x$, follow these steps:

Step 1: Rewrite $3^{2x}$ using properties of exponents

We know that: $3^{2x} = (3^x)^2$ So, the equation becomes: $(3^x)^2 + 9 = 10 \cdot 3^x$

Step 2: Substitute $y = 3^x$

This substitution simplifies the equation to: $y^2 + 9 = 10y$

Step 3: Rearrange into standard quadratic form

$y^2 - 10y + 9 = 0$

Step 4: Solve the quadratic equation

Use the quadratic formula: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = -10$, $c = 9$. Substituting these values: $y = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(9)}}{2(1)}$ $y = \frac{10 \pm \sqrt{100 - 36}}{2}$ $y = \frac{10 \pm \sqrt{64}}{2}$ $y = \frac{10 \pm 8}{2}$

Step 5: Calculate the two solutions for $y$

$y = \frac{10 + 8}{2} = 9 \quad \text{and} \quad y = \frac{10 - 8}{2} = 1$

Step 6: Back-substitute $y = 3^x$

1. For $y = 9$: $3^x = 9 \quad \implies \quad x = 2$
2. For $y = 1$: $3^x = 1 \quad \implies \quad x = 0$

Final Solution:

The solutions are: $x = 2 \quad \text{and} \quad x = 0$

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Would you like further details or have additional questions?

Here are 5 related questions to expand on this topic:

1. How do we solve equations involving exponents with different bases?
2. What are some real-world applications of exponential equations?
3. How do we handle cases where the quadratic equation has no real solutions?
4. Can this method be applied to equations with more complex exponents?
5. What is the significance of substitution in solving exponential equations?

Tip: Always check your solutions by substituting them back into the original equation!