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To solve the equation \(6^{2x} + 6^x - 12 = 0\), we can simplify it by using substitution.

### Step-by-Step Solution:
1. **Substitute**: Let \(y = 6^x\). Then \(6^{2x}\) can be rewritten as \((6^x)^2 = y^2\).
   
   This transforms the equation into:
   \[
   y^2 + y - 12 = 0
   \]

2. **Factor the Equation**: We now have a quadratic equation in terms of \(y\):
   \[
   y^2 + y - 12 = 0
   \]
   To factor it, we look for two numbers that multiply to \(-12\) and add up to \(1\). These numbers are \(4\) and \(-3\).
   
   Thus,
   \[
   (y + 4)(y - 3) = 0
   \]

3. **Solve for \(y\)**: Set each factor equal to zero:
   \[
   y + 4 = 0 \quad \Rightarrow \quad y = -4
   \]
   \[
   y - 3 = 0 \quad \Rightarrow \quad y = 3
   \]

   Since \(y = 6^x\) represents an exponential function and cannot be negative, we discard \(y = -4\).

   So, we only have:
   \[
   y = 3
   \]

4. **Back-Substitute for \(x\)**:
   \[
   6^x = 3
   \]
   To solve for \(x\), take the logarithm of both sides:
   \[
   x = \frac{\ln(3)}{\ln(6)}
   \]

5. **Solution**:
   - **Exact form**: \( x = \frac{\ln(3)}{\ln(6)} \)
   - **Decimal form**: Calculating the value, \( x \approx 0.613\) (rounded to three decimal places).

### Final Answer:
- The solution set is: \( \left\{ \frac{\ln(3)}{\ln(6)} \right\} \).

Would you like further details or have any questions?

### Related Questions:
1. How would you solve a similar equation if the base was different, like \(4^{2x} + 4^x - 12 = 0\)?
2. What if the equation was \(6^{2x} - 6^x + 12 = 0\)?
3. How does taking logarithms help in solving exponential equations?
4. Can this method of substitution be applied to other types of equations?
5. How do you handle complex solutions in exponential equations?

### Tip:
When dealing with exponential equations, substitution is a useful method for simplifying the equation, especially when terms appear in powers of the same base.

Solve the following exponential equation. Express irrational solutions in exact form and as a decimal rounded to three decimal places. 6^(2x) + 6^x - 12 = 0

Learn how to solve the exponential equation 6^(2x) + 6^x - 12 = 0 using substitution and logarithmic conversion. This solution provides exact and decimal forms, suitable for high school students in Grades 10-12.

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