Solve the equation 8(82^x)+39(8^x)−5=08(82^x)+39(8^x)−5=0 for xx.

The equation to solve is:

\[
8(82^x) + 39(8^x) - 5 = 0
\]

Let's solve this equation step by step.

### 1. **Substitute** 
Let \( y = 8^x \). Then, the equation becomes:

\[
8(82^x) + 39(8^x) - 5 = 8 \cdot (8^x)^2 + 39 \cdot 8^x - 5 = 0
\]

This can be written as a quadratic equation:

\[
8y^2 + 39y - 5 = 0
\]

### 2. **Solve the quadratic equation**
We use the quadratic formula:

\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For our equation \( 8y^2 + 39y - 5 = 0 \):
- \( a = 8 \)
- \( b = 39 \)
- \( c = -5 \)

Substituting these values:

\[
y = \frac{-39 \pm \sqrt{39^2 - 4(8)(-5)}}{2(8)}
\]

\[
y = \frac{-39 \pm \sqrt{1521 + 160}}{16}
\]

\[
y = \frac{-39 \pm \sqrt{1681}}{16}
\]

\[
y = \frac{-39 \pm 41}{16}
\]

This gives two solutions for \( y \):

\[
y = \frac{2}{16} = \frac{1}{8} \quad \text{or} \quad y = \frac{-80}{16} = -5
\]

### 3. **Back-substitute \( y = 8^x \)**
Since \( y = 8^x \):

1. For \( 8^x = \frac{1}{8} \):
   \[
   8^x = 8^{-1} \implies x = -1
   \]

2. For \( 8^x = -5 \):
   This has no real solution since \( 8^x > 0 \) for all real \( x \).

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

Would you like more details or have any questions?

### Related Questions
1. How would the solution change if the equation was \(8(82^x) + 39(8^x) + 5 = 0\)?
2. How do you verify the solution \(x = -1\)?
3. What would happen if the base were different, like \(9^x\) instead of \(8^x\)?
4. Can there be a complex solution for \(8^x = -5\)?
5. What are the properties of exponential functions that lead to \(8^x > 0\)?

### Tip
When solving equations involving exponential terms, consider using substitutions to reduce the equation to a more familiar form, such as a quadratic.

Learn how to solve the equation 8(82^x) + 39(8^x) - 5 = 0 for x step-by-step with detailed explanations and solutions. This problem involves exponential functions and quadratic equations, suitable for advanced high school students.

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