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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.

Nilai dari log_3(9) + log_3(108) - log_9(4) + log_9(27) dan beberapa masalah logaritma terkait.

Solve logarithmic expressions involving different bases, using properties of logarithms like the change of base formula. Suitable for students in Grades 10-12. Includes step-by-step breakdown of log_3(9) + log_3(108) - log_9(4) + log_9(27) and other logarithmic problems.

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