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To solve the exponential equation

\[
7^{2x} + 7^x - 6 = 0,
\]

we can simplify this by substituting \( y = 7^x \), which turns the equation into a quadratic form:

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

Now, we solve this quadratic equation for \( y \).

### Step 1: Factor the Quadratic Equation
We need two numbers that multiply to \(-6\) and add to \(1\). These numbers are \(3\) and \(-2\), so we can factor the equation as follows:

\[
(y + 3)(y - 2) = 0.
\]

### Step 2: Solve for \( y \)
This gives us two possible values for \( y \):

1. \( y + 3 = 0 \Rightarrow y = -3 \)
2. \( y - 2 = 0 \Rightarrow y = 2 \)

Since \( y = 7^x \) and \(7^x\) is always positive, we discard \( y = -3 \) because it has no real solution in the context of exponentials.

Thus, we are left with:

\[
7^x = 2.
\]

### Step 3: Solve for \( x \)
To solve \(7^x = 2\), we take the logarithm of both sides (using natural logarithms for convenience):

\[
x \ln 7 = \ln 2.
\]

Then we solve for \( x \):

\[
x = \frac{\ln 2}{\ln 7}.
\]

### Step 4: Approximate the Solution
Using a calculator, we find:

\[
x \approx \frac{\ln 2}{\ln 7} \approx 0.356.
\]

### Final Answer
The solution set is:

\[
\{ x \approx 0.356 \}.
\]

So the correct choice is:

**A. The solution set is \(\{ 0.356 \}\),** rounded to three decimal places.

Would you like further details or have any questions?

Here are some related questions to explore:

1. How can we verify if \( x = 0.356 \) is a solution to the original equation?
2. What is the general approach for solving exponential equations that are quadratic in form?
3. Why do we discard \( y = -3 \) as a solution for \( 7^x = -3 \)?
4. Can this method be used for any base other than \( 7 \) in a similar equation?
5. How would the solution change if the original equation was \( 7^{2x} - 7^x - 6 = 0 \)?

**Tip:** When solving exponential equations, a substitution can often simplify the equation into a form that is easier to work with, such as a quadratic equation.

Solve the following exponential equation. Express irrational solutions in exact form and as a decimal rounded to three decimal places. 7^(2x) + 7^x - 6 = 0.

Learn how to solve the exponential equation 7^(2x) + 7^x - 6 = 0 using substitution and logarithmic properties. This solution expresses the answer in exact form and as a decimal rounded to three decimal places. Ideal for high school students in Grades 10-12.

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