The question you are asking about is part (c) in the image:

$\text{If } x > 0 \text{ and } x \neq 1, \log 6 = a \text{ and } \log 8 = b, \text{ determine the value of } 3a - b.$

Step-by-Step Solution:

We are given the following information:

• $\log 6 = a$
• $\log 8 = b$
• $x > 0$ and $x \neq 1$

We are tasked with finding the value of $3a - b$.

1. Express $a$ in terms of logarithms:

$\log 6 = a$ Using the logarithmic property $\log(ab) = \log a + \log b$, we can break down $\log 6$ as: $a = \log (2 \cdot 3) = \log 2 + \log 3$ So, $a = \log 2 + \log 3$.

2. Express $b$ in terms of logarithms:

$\log 8 = b$ Since $8 = 2^3$, we use the power rule $\log a^n = n \log a$: $b = \log (2^3) = 3 \log 2$

3. Substitute into the expression for $3a - b$:

Now, substitute the expressions for $a$ and $b$ into $3a - b$: $3a - b = 3(\log 2 + \log 3) - 3 \log 2$

4. Simplify the expression:

Expand the terms: $3a = 3(\log 2 + \log 3) = 3 \log 2 + 3 \log 3$ Now subtract $b = 3 \log 2$: $3a - b = (3 \log 2 + 3 \log 3) - 3 \log 2$ The $3 \log 2$'s cancel out, leaving: $3a - b = 3 \log 3$

Final Answer:

$3a - b = 3 \log 3$

This is the simplified value of $3a - b$.

Would you like any further clarifications?

Here are 5 related questions:

1. How can the properties of logarithms simplify expressions with products and powers?
2. How does the cancellation of terms work when simplifying logarithms?
3. How do logarithms help in solving exponential equations?
4. What happens when we change the base of logarithms in a problem like this?
5. Can the solution method be extended to different logarithmic expressions involving more terms?

Tip: When working with logarithmic equations, always try to break down the logarithms using properties like the product, quotient, and power rules to simplify the terms.