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### Problem 18: Equation of the line passing through \((-1, 7)\) and parallel to the line through \((-3, -5)\) and \((1, 3)\).

We need the equation of the line that passes through the point \((-1, 7)\) and is parallel to the line passing through \((-3, -5)\) and \((1, 3)\).

1. First, find the slope of the line passing through the points \((-3, -5)\) and \((1, 3)\). The slope formula is:

   \[
   m = \frac{y_2 - y_1}{x_2 - x_1}
   \]

   Substituting the points \((-3, -5)\) and \((1, 3)\):

   \[
   m = \frac{3 - (-5)}{1 - (-3)} = \frac{3 + 5}{1 + 3} = \frac{8}{4} = 2
   \]

   So, the slope of the line is \(m = 2\).

2. Since the line we want is parallel to this line, its slope is also \(m = 2\).

3. Now, use the point-slope form of the equation of a line:

   \[
   y - y_1 = m(x - x_1)
   \]

   Substituting \(m = 2\) and the point \((-1, 7)\):

   \[
   y - 7 = 2(x - (-1)) = 2(x + 1)
   \]

   Simplifying:

   \[
   y - 7 = 2x + 2 \implies y = 2x + 9
   \]

   Therefore, the equation of the line is:

   \[
   y = 2x + 9
   \]

### Problem 19: Solve the equation \(8^x = \left(\frac{1}{128}\right)^{x - 8}\).

1. Rewrite \(128\) as a power of 2: \(128 = 2^7\), so \(\frac{1}{128} = 2^{-7}\).

   The equation becomes:

   \[
   8^x = (2^{-7})^{x - 8}
   \]

2. Express \(8\) as a power of 2: \(8 = 2^3\). So the equation is:

   \[
   (2^3)^x = 2^{-7(x - 8)}
   \]

   Simplifying:

   \[
   2^{3x} = 2^{-7(x - 8)}
   \]

3. Since the bases are the same, set the exponents equal:

   \[
   3x = -7(x - 8)
   \]

4. Distribute the \(-7\) on the right:

   \[
   3x = -7x + 56
   \]

5. Solve for \(x\):

   \[
   3x + 7x = 56 \implies 10x = 56 \implies x = \frac{56}{10} = 5.6
   \]

   Therefore, the solution is \(x = 5.6\).

### Problem 20: Find the domain of the function \(g(x) = 6(x - 1)^{5/2}\).

The domain of a function involving a radical must ensure the expression inside the radical is non-negative (since the even root of a negative number is not a real number). Here, we have \((x - 1)^{5/2}\).

1. For the square root to be real, \(x - 1 \geq 0\).

2. Solving for \(x\):

   \[
   x - 1 \geq 0 \implies x \geq 1
   \]

   So, the domain is \(x \in [1, \infty)\).

### Problem 21: Solve the equation \(5x^2 - 10 \cdot 5^x + 25 = 0\).

Let’s solve this step by step.

1. First, simplify the equation by observing that \(5^x\) appears in two terms. Let \(y = 5^x\), so the equation becomes:

   \[
   5y^2 - 10y + 25 = 0
   \]

2. Solve this quadratic equation using the quadratic formula:

   \[
   y = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(5)(25)}}{2(5)}
   \]

   Simplifying:

   \[
   y = \frac{10 \pm \sqrt{100 - 500}}{10} = \frac{10 \pm \sqrt{-400}}{10}
   \]

   Since the discriminant is negative, this equation has no real solutions.

   Therefore, the original equation has no real solutions.

---

Do you need more details or explanations? 

Here are 5 questions that are related:

1. How do you find the slope of a line between two points?
2. What are the conditions for parallel lines in terms of slope?
3. How can you rewrite exponential expressions with different bases?
4. What is the general method for finding the domain of a function?
5. How can you solve a quadratic equation using substitution?

**Tip**: Always check the discriminant in a quadratic equation to determine if it has real, complex, or no solutions.

Find an equation of the line that passes through the point (-1, 7) and is parallel to the line passing through the points (-3, -5) and (1, 3). Solve the equation for x: 8^x = (1/128)^(x-8). Find the domain of the function g(x) = 6(x-1)^(5/2). Solve the equation for x: 5x^2 - 10*5^x + 25 = 0.

Learn how to find the equation of a line parallel to another, solve exponential equations like 8^x = (1/128)^(x-8), find the domain of a radical function, and solve quadratic equations. This tutorial is suitable for students in Grades 9-11, covering key algebraic and geometric concepts with step-by-step explanations.

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