Let's solve the equation \(2^{(x+3)} = 2^{(1-x)} + 15\).

### Step 1: Simplify the equation by isolating the exponential terms
We have:
\[
2^{(x+3)} = 2^{(1-x)} + 15
\]
Subtract \(2^{(1-x)}\) from both sides to get:
\[
2^{(x+3)} - 2^{(1-x)} = 15
\]

### Step 2: Express the exponents in a simpler form
Rewrite \(2^{(x+3)}\) and \(2^{(1-x)}\):
\[
2^{(x+3)} = 2^x \cdot 2^3 = 8 \cdot 2^x
\]
\[
2^{(1-x)} = \frac{2}{2^x}
\]
So, the equation becomes:
\[
8 \cdot 2^x - \frac{2}{2^x} = 15
\]

### Step 3: Multiply through by \(2^x\) to eliminate the fraction
Multiply every term by \(2^x\) to get:
\[
8 \cdot (2^x)^2 - 2 = 15 \cdot 2^x
\]

### Step 4: Introduce a substitution
Let \(y = 2^x\). The equation now is:
\[
8y^2 - 15y - 2 = 0
\]

### Step 5: Solve the quadratic equation
Use the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 8\), \(b = -15\), and \(c = -2\):
\[
y = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 8 \cdot (-2)}}{16}
\]
\[
y = \frac{15 \pm \sqrt{225 + 64}}{16}
\]
\[
y = \frac{15 \pm \sqrt{289}}{16}
\]
\[
y = \frac{15 \pm 17}{16}
\]

This gives two possible values for \(y\):
\[
y = \frac{32}{16} = 2 \quad \text{or} \quad y = \frac{-2}{16} = -\frac{1}{8}
\]

### Step 6: Solve for \(x\)
Recall that \(y = 2^x\). Therefore:
- If \(y = 2\), then \(2^x = 2\), so \(x = 1\).
- If \(y = -\frac{1}{8}\), there's no real solution because \(2^x\) is always positive.

### Final Answer
The only solution is \(x = 1\).

Would you like more details or have any questions?

Here are 5 related questions you might find interesting:
1. What is the general method for solving exponential equations?
2. How do you solve equations involving logarithms?
3. Can you solve \(3^{x+2} = 5 \cdot 3^{x-1} + 18\)?
4. What happens when you have an exponential equation with different bases?
5. How does the quadratic formula apply to different types of equations?

**Tip:** Always check for extraneous solutions, especially when dealing with exponential and logarithmic equations.

Learn how to solve the exponential equation 2^(x+3) = 2^(1-x) + 15 using properties of exponents and quadratic equations. This problem provides a detailed step-by-step solution, covering key algebraic concepts and applying the quadratic formula. Suitable for students in Grades 9-12.

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