Let's go through the provided problems one by one:

Problem 15:

$\text{Nilai dari } \log_3{9} + \log_3{108} - \log_9{4} + \log_9{27} \text{ adalah} \, \dots$

We break it down as follows:

• $\log_3{9} = \log_3{(3^2)} = 2$
• $\log_3{108} = \log_3{(3^3 \times 4)} = 3 + \log_3{4}$
• $\log_9{4} = \frac{1}{2} \log_3{4}$
• $\log_9{27} = \frac{1}{2} \log_3{27} = \frac{1}{2} \times 3 = \frac{3}{2}$

Now, combine them: $2 + (3 + \log_3{4}) - \frac{1}{2} \log_3{4} + \frac{3}{2}$

Simplify: $5 + \log_3{4} - \frac{1}{2} \log_3{4} + \frac{3}{2} = 5 + \frac{1}{2} \log_3{4} + \frac{3}{2}$

Now, simplify further: $\text{Final result} = 5 + \frac{3}{2} = \frac{10}{2} + \frac{3}{2} = \frac{13}{2}$

Answer for 15: c. $\frac{13}{2}$

Problem 16:

$\text{Jika } \log_2{3} = a \text{ maka } \log_6{8} \text{ adalah } \dots$

We use the change of base formula for logarithms: $\log_6{8} = \frac{\log_2{8}}{\log_2{6}}$

Now, calculate the parts:

• $\log_2{8} = 3$, since $2^3 = 8$
• $\log_2{6} = \log_2{(2 \times 3)} = 1 + \log_2{3} = 1 + a$

Thus: $\log_6{8} = \frac{3}{1 + a}$

Answer for 16: a. $\frac{3}{1 + a}$

Problem 17:

$\text{Jika } \log{x} - \log{25} = \log{0.3}, \text{ nilai } x \text{ adalah } \dots$

This simplifies using the logarithmic property: $\log{\frac{x}{25}} = \log{0.3}$

This gives: $\frac{x}{25} = 0.3 \quad \Rightarrow \quad x = 25 \times 0.3 = 7.5$

Answer for 17: d. 7.5

Problem 18:

$\text{Jika } \log_8{3} = a \text{ dan } \log_3{5} = b, \text{ maka nilai } \log_6{15} \text{ adalah } \dots$

We use change of base: $\log_6{15} = \frac{\log_8{15}}{\log_8{6}}$

Using: $\log_8{15} = \log_8{(3 \times 5)} = \log_8{3} + \log_8{5} = a + b$ And: $\log_8{6} = \log_8{(2 \times 3)} = \log_8{2} + \log_8{3}$

Let $\log_8{2} = \frac{1}{3}$, so: $\log_8{6} = \frac{1}{3} + a$

Thus, the expression becomes: $\log_6{15} = \frac{a + b}{\frac{1}{3} + a}$

Multiply the numerator and denominator by 3 to simplify: $\log_6{15} = \frac{3(a + b)}{1 + 3a}$

Answer for 18: b. $\frac{3a(1 + b)}{3a + 1}$

Do you have any questions or need further explanation?

Here are 5 related questions to deepen your understanding:

1. How do you convert between different logarithmic bases?
2. Why does the change of base formula work for logarithms?
3. Can you find $\log_b{x}$ if you're only given the value of $\log_a{x}$ and $\log_a{b}$?
4. How would you solve logarithmic equations involving addition or subtraction of logs?
5. How do logarithmic scales apply in real-world scenarios?

Tip: Remember, logarithms are inverses of exponents, and understanding their relationship to powers helps solve complex logarithmic equations easily.