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.