Let's solve the equations step by step as per the instructions in the image.

1. Solve the quadratic equation:

$y^2 - 7y - 8 = 0$

Step 1: Identify the coefficients.

The quadratic equation is in the form $ay^2 + by + c = 0$, where:

• $a = 1$
• $b = -7$
• $c = -8$

Step 2: Use the quadratic formula:

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

Substitute the values of $a$, $b$, and $c$: $y = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(-8)}}{2(1)}$ $y = \frac{7 \pm \sqrt{49 + 32}}{2}$ $y = \frac{7 \pm \sqrt{81}}{2}$ $y = \frac{7 \pm 9}{2}$

Step 3: Calculate the roots:

1. $y_1 = \frac{7 + 9}{2} = \frac{16}{2} = 8$
2. $y_2 = \frac{7 - 9}{2} = \frac{-2}{2} = -1$

Solution: $y = 8$ or $y = -1$.

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2. Solve the logarithmic equation:

$\log(x - 1) - \log(x + 3) = \log 2$

Step 1: Combine the logarithms using the property:

$\log\left(\frac{x - 1}{x + 3}\right) = \log 2$

Step 2: Remove the logarithms by exponentiating both sides:

$\frac{x - 1}{x + 3} = 2$

Step 3: Solve the resulting equation:

$x - 1 = 2(x + 3)$ $x - 1 = 2x + 6$ $x - 2x = 6 + 1$ $-x = 7$ $x = -7$

Solution: $x = -7$.

However, this solution needs to be checked against the original equation to ensure it doesn't create a negative argument in the logarithm. Since $x = -7$ makes $\log(x - 1)$ and $\log(x + 3)$ undefined, there is no valid solution for this equation.

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3. Solve the exponential equation:

$2 \cdot 3^x + 6 \cdot 3^{-x} = 1$

Step 1: Set $y = 3^x$, then the equation becomes:

$2y + \frac{6}{y} = 1$

Step 2: Multiply the entire equation by $y$ to eliminate the fraction:

$2y^2 + 6 = y$

Step 3: Rearrange to form a quadratic equation:

$2y^2 - y + 6 = 0$

Step 4: Solve the quadratic equation using the quadratic formula:

$y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(6)}}{2(2)}$ $y = \frac{1 \pm \sqrt{1 - 48}}{4}$ $y = \frac{1 \pm \sqrt{-47}}{4}$

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

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Summary of Solutions:

1. $y = 8$ or $y = -1$
2. No valid solution for the logarithmic equation.
3. No real solutions for the exponential equation.

Would you like more details on any of these solutions or need further assistance?

Here are 5 related questions:

1. How can the discriminant help in identifying the nature of the roots of a quadratic equation?
2. What are the conditions under which a logarithmic equation has no solution?
3. What are complex numbers, and how are they used in solving quadratic equations with negative discriminants?
4. Can all exponential equations be solved using substitution? Why or why not?
5. What are some practical applications of solving quadratic equations?

Tip: Always check the domain of the functions involved (like logarithms and square roots) when solving equations to ensure the solutions are valid.