The equation given is:

\[
10^{2x} - 8 \cdot 10^x + 12 = 0
\]

This is a quadratic equation in terms of \( 10^x \). To simplify the equation, let \( y = 10^x \). Therefore, the equation becomes:

\[
y^2 - 8y + 12 = 0
\]

Now, we can solve this quadratic equation using the quadratic formula:

\[
y = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(12)}}{2(1)}
\]

Simplifying:

\[
y = \frac{8 \pm \sqrt{64 - 48}}{2}
\]
\[
y = \frac{8 \pm \sqrt{16}}{2}
\]
\[
y = \frac{8 \pm 4}{2}
\]

Thus, the two possible values of \( y \) are:

\[
y = \frac{8 + 4}{2} = 6 \quad \text{or} \quad y = \frac{8 - 4}{2} = 2
\]

Now, recall that \( y = 10^x \). Therefore:

1. \( 10^x = 6 \)
2. \( 10^x = 2 \)

To solve for \( x \), we take the logarithm of both sides:

1. \( x = \log_{10}(6) \approx 0.778 \)
2. \( x = \log_{10}(2) \approx 0.301 \)

So, the solutions are:

\[
x \approx 0.778 \quad \text{or} \quad x \approx 0.301
\]

Would you like more details or clarifications?

Here are some related questions:
1. What are the steps to solving general quadratic equations?
2. How does the substitution \( y = 10^x \) simplify exponential equations?
3. What is the quadratic formula, and how is it derived?
4. How can logarithms be used to solve exponential equations?
5. Can there be more complex solutions for similar exponential equations?

**Tip**: Whenever solving exponential equations, look for ways to reduce them into simpler forms, such as quadratic equations, to make solving easier.

Learn how to solve the exponential equation 10^2x−8⋅10^x+12 by converting it into a quadratic equation using substitution and the quadratic formula. This solution demonstrates the use of logarithms and key algebraic methods, suitable for students in Grades 9-12.

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