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The equation given is:

\[
3^{2x} + 3^x - 6 = 0
\]

To solve for \(x\), we can simplify this by making a substitution. Let \(y = 3^x\). Then, \(3^{2x} = (3^x)^2 = y^2\). Substituting into the equation gives:

\[
y^2 + y - 6 = 0
\]

This is a quadratic equation in \(y\). We can solve it using the quadratic formula:

\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \(a = 1\), \(b = 1\), and \(c = -6\). Substituting these values into the quadratic formula:

\[
y = \frac{-1 \pm \sqrt{1^2 - 4(1)(-6)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm \sqrt{25}}{2}
\]

\[
y = \frac{-1 \pm 5}{2}
\]

This gives two solutions for \(y\):

\[
y = \frac{-1 + 5}{2} = 2 \quad \text{or} \quad y = \frac{-1 - 5}{2} = -3
\]

Since \(y = 3^x\) and \(y = 2\), we solve for \(x\):

\[
3^x = 2 \quad \Rightarrow \quad x = \log_3(2)
\]

For the case where \(y = -3\), since \(3^x\) is always positive for real \(x\), there is no solution from this case.

Thus, the only solution is:

\[
x = \log_3(2)
\]

Would you like a more detailed explanation of logarithms or solving similar equations?

Here are 5 questions to expand on this topic:
1. What are the steps to convert between logarithmic and exponential forms?
2. How would you solve an equation like \(2^{2x} + 2^x - 8 = 0\)?
3. What is the meaning of \(\log_3(2)\), and how is it approximated?
4. How can the quadratic formula be applied to other types of functions?
5. Are there any methods to solve exponential equations without substitution?

**Tip**: Always check the domain of functions when solving equations to make sure solutions are valid.

Solve for x in the equation 3^(2x) + 3^x - 6 = 0.

Learn how to solve the exponential equation 3^(2x) + 3^x - 6 = 0 using substitution and the quadratic formula. This step-by-step guide covers key concepts in algebra and is suitable for students in Grades 9-12.

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