https://file.aiquickdraw.com/mathbot/file/172663052107548dypv34.jpg

The equation in the image is:

\[
4^y - 3(2^y) - 10 = 0
\]

We can solve this equation step-by-step:

### Step 1: Substitute \(2^y = x\)
This simplifies the equation into a quadratic form:

\[
4^y = (2^2)^y = (2^y)^2 = x^2
\]

Thus, the equation becomes:

\[
x^2 - 3x - 10 = 0
\]

### Step 2: Solve the quadratic equation
We can solve the quadratic equation \(x^2 - 3x - 10 = 0\) using the quadratic formula:

\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-10)}}{2(1)}
\]

\[
x = \frac{3 \pm \sqrt{9 + 40}}{2}
\]

\[
x = \frac{3 \pm \sqrt{49}}{2}
\]

\[
x = \frac{3 \pm 7}{2}
\]

So, the two solutions for \(x\) are:

\[
x = \frac{3 + 7}{2} = 5 \quad \text{or} \quad x = \frac{3 - 7}{2} = -2
\]

### Step 3: Back-substitute for \(x = 2^y\)
- If \(x = 5\), then \(2^y = 5\), so \(y = \log_2(5)\).
- If \(x = -2\), this does not have a real solution because \(2^y = -2\) is impossible (as powers of 2 are always positive).

Thus, the only valid solution is:

\[
y = \log_2(5)
\]

This matches option **a** in the answer choices. Therefore, the correct answer is **A**.

---

Do you want more details or have any questions? Here are 5 related questions to deepen your understanding:

1. How do you solve quadratic equations using factorization?
2. What is the quadratic formula, and when is it used?
3. How can logarithms help in solving exponential equations?
4. Why are negative powers of 2 impossible in real numbers?
5. How does the change of base formula work for logarithms?

**Tip**: Always check if quadratic equations can be factored before applying the quadratic formula—it may save time!

Solve the equation: 4^y - 3(2^y) - 10 = 0.

Learn how to solve the exponential equation 4^y - 3(2^y) - 10 = 0 by transforming it into a quadratic equation and using logarithms. This step-by-step guide covers quadratic equations, logarithmic functions, and the quadratic formula, suitable for Grades 10-12 students.

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