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Let's solve both equations provided in parts (b) and (c).

### Problem (b)
Given:
\[
8^x - 8^{-x} = 8
\]

To solve this equation, let’s try using substitution to simplify the expression. Set:
\[
y = 8^x
\]
Then, \(8^{-x} = \frac{1}{y}\), so the equation becomes:
\[
y - \frac{1}{y} = 8
\]
Multiply through by \(y\) to eliminate the fraction:
\[
y^2 - 1 = 8y
\]
Rearrange this to form a quadratic equation:
\[
y^2 - 8y - 1 = 0
\]
Now, solve this quadratic equation using the quadratic formula:
\[
y = \frac{8 \pm \sqrt{64 + 4}}{2} = \frac{8 \pm \sqrt{68}}{2} = \frac{8 \pm 2\sqrt{17}}{2} = 4 \pm \sqrt{17}
\]
So we have:
\[
y = 4 + \sqrt{17} \quad \text{or} \quad y = 4 - \sqrt{17}
\]

Since \(y = 8^x\) must be positive, only \(y = 4 + \sqrt{17}\) is valid.

Now, solve for \(x\):
\[
8^x = 4 + \sqrt{17}
\]
Take the logarithm of both sides:
\[
x \ln 8 = \ln (4 + \sqrt{17})
\]
\[
x = \frac{\ln (4 + \sqrt{17})}{\ln 8}
\]

### Problem (c)
Given:
\[
\frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{1}{2}
\]

To solve this, let’s use the hyperbolic identity by setting:
\[
y = e^x
\]
Then, \(e^{-x} = \frac{1}{y}\), and the equation becomes:
\[
\frac{y - \frac{1}{y}}{y + \frac{1}{y}} = \frac{1}{2}
\]
Multiply both the numerator and denominator by \(y\):
\[
\frac{y^2 - 1}{y^2 + 1} = \frac{1}{2}
\]
Cross-multiply to clear the fraction:
\[
2(y^2 - 1) = y^2 + 1
\]
Expanding and simplifying:
\[
2y^2 - 2 = y^2 + 1
\]
\[
y^2 = 3
\]
\[
y = \pm \sqrt{3}
\]

Since \(y = e^x\), and \(y\) must be positive:
\[
e^x = \sqrt{3}
\]
Take the natural logarithm of both sides:
\[
x = \ln \sqrt{3} = \frac{\ln 3}{2}
\]

### Summary of Solutions:
- **(b)** \( x = \frac{\ln (4 + \sqrt{17})}{\ln 8} \)
- **(c)** \( x = \frac{\ln 3}{2} \)

Would you like further details on any of these steps?

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### Related Questions
1. How can logarithmic properties help solve exponential equations?
2. What other techniques can be used for solving quadratic equations?
3. How does substitution simplify solving exponential equations?
4. What are the properties of hyperbolic functions, and how do they relate to exponential functions?
5. How does the natural logarithm simplify expressions involving \(e\)?

### Tip
Using substitution can simplify complex expressions and make solving exponential or logarithmic equations easier.

Solve the equations: (b) 8^x - 8^(-x) = 8, (c) (e^x - e^(-x)) / (e^x + e^(-x)) = 1/2.

Explore how to solve exponential equations like 8^x - 8^(-x) = 8 and (e^x - e^(-x)) / (e^x + e^(-x)) = 1/2 using logarithms, quadratic equations, and hyperbolic identities. Ideal for advanced math students or college-level learners.

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