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Let's go over each equation and solve them one by one.

### (a) \( e^x - e^{-x} = 5 \)
This equation resembles a hyperbolic sine function:
\[
e^x - e^{-x} = 2\sinh(x)
\]
Thus, the equation can be rewritten as:
\[
2\sinh(x) = 5 \quad \Rightarrow \quad \sinh(x) = \frac{5}{2}
\]
Taking the inverse hyperbolic sine (arsinh) on both sides:
\[
x = \sinh^{-1}\left(\frac{5}{2}\right)
\]
Using a calculator:
\[
x \approx 1.4436
\]

### (b) \( e^x + e^{-x} = 3 \)
This equation resembles a hyperbolic cosine function:
\[
e^x + e^{-x} = 2\cosh(x)
\]
Thus, the equation can be rewritten as:
\[
2\cosh(x) = 3 \quad \Rightarrow \quad \cosh(x) = \frac{3}{2}
\]
Taking the inverse hyperbolic cosine (arcosh) on both sides:
\[
x = \cosh^{-1}\left(\frac{3}{2}\right)
\]
Using a calculator:
\[
x \approx 0.9624
\]

### (c) \( \ln(x^2) + \ln(2x^4) = 16 \)
Using the logarithmic properties, we can combine the terms:
\[
\ln(x^2) + \ln(2x^4) = \ln(x^2 \cdot 2x^4) = \ln(2x^6)
\]
Thus, the equation becomes:
\[
\ln(2x^6) = 16
\]
Exponentiating both sides:
\[
2x^6 = e^{16}
\]
Solving for \( x^6 \):
\[
x^6 = \frac{e^{16}}{2}
\]
Taking the 6th root:
\[
x = \left(\frac{e^{16}}{2}\right)^{1/6}
\]
Approximating using a calculator:
\[
x \approx 6.448
\]

### (d) \( 6^{-x} = 20 \)
Taking the natural logarithm of both sides:
\[
\ln(6^{-x}) = \ln(20) \quad \Rightarrow \quad -x \ln(6) = \ln(20)
\]
Solving for \( x \):
\[
x = -\frac{\ln(20)}{\ln(6)}
\]
Using a calculator:
\[
x \approx -1.5849
\]

---

If you'd like more details or have any questions about any of the steps, feel free to ask! Here are 5 related questions that may expand your understanding:

1. How are hyperbolic functions related to exponential functions?
2. What are the properties of logarithms used to simplify equation (c)?
3. How can you solve general exponential equations like \( a^x = b \)?
4. What is the difference between inverse hyperbolic functions and their trigonometric counterparts?
5. How would you handle solving an equation like \( e^x - e^{-x} = k \) for arbitrary \( k \)?

**Tip:** Always check if an equation involving exponentials can be expressed in terms of hyperbolic functions to simplify the solving process.

Solve the following equations: (a) e^x - e^(-x) = 5, (b) e^x + e^(-x) = 3, (c) ln(x^2) + ln(2x^4) = 16, (d) 6^(-x) = 20

Explore solutions to exponential and logarithmic equations involving hyperbolic functions. Detailed steps include solving equations like e^x - e^(-x) = 5 and ln(x^2) + ln(2x^4) = 16, with essential formulas and theorems. Suitable for advanced students.

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