The image contains several logarithmic exercises in Indonesian. I'll help break down and solve them. Here are the questions and their solutions:

1. Dengan menggunakan sifat logaritma, tentukan nilai berikut:

(Use the properties of logarithms to determine the values.)

a. $\log 64$

Solution: $64 = 2^6$ By the logarithmic property $\log_b(a^n) = n \log_b a$, $\log 64 = \log (2^6) = 6 \log 2$ Thus, $\log 64$ depends on $\log 2$. If $\log 2$ is known, this can be computed, or this is the simplified form.

b. $\log 216$

Solution: $216 = 6^3$ By the same logarithmic property, $\log 216 = \log (6^3) = 3 \log 6$ Thus, $\log 216 = 3 \log 6$.

c. $\log 0.0001$

Solution: $0.0001 = 10^{-4}$ So, $\log 0.0001 = \log (10^{-4}) = -4$

2. Sederhanakanlah. (Simplify the following)

a. $\log_2 320 + \log_3 27 - \log_2 27 - \log_5 625$

Solution:

• $\log_3 27 = \log_3(3^3) = 3$
• $\log_5 625 = \log_5(5^4) = 4$

This equation depends on the base for $\log_2 320$ and $\log_2 27$. These would typically require a calculator if the exact values are needed, but the rest simplifies as: $= \log_2 320 + 3 - \log_2 27 - 4$

b. $\log_5 200 + \log_5 16 - \log_5 16 - \log_5 8$

Simplify:

• $\log_5 16 - \log_5 16 = 0$ (these terms cancel out) So, $\log_5 200 - \log_5 8$ Apply the property $\log_b a - \log_b c = \log_b \left(\frac{a}{c}\right)$, $= \log_5 \left(\frac{200}{8}\right) = \log_5 25$ Since $25 = 5^2$, $\log_5 25 = 2$

3. Jika diketahui $\log 3 = 0.4771$ dan $\log 5 = 0.6990$, tentukan nilai berikut.

(Given $\log 3 = 0.4771$ and $\log 5 = 0.6990$, determine the values.)

a. $\log 45$

Solution: $\log 45 = \log(5 \times 9) = \log 5 + \log 9$ Since $\log 9 = \log(3^2) = 2 \log 3$, $\log 45 = \log 5 + 2 \log 3 = 0.6990 + 2(0.4771) = 0.6990 + 0.9542 = 1.6532$

b. $\log 25$

Solution: $\log 25 = \log(5^2) = 2 \log 5 = 2(0.6990) = 1.398$

4. Tentukan nilai $\log_2^2 \log_3^3 \log_5^5 \log_6^6 \log_8^8$

Solution: To compute the value of this expression, you would apply logarithmic properties. This seems to be an advanced expression that combines powers of logs. Without further context or simplifications, it's hard to proceed, but this requires careful application of logarithmic rules.

Let me know if you need further clarification on any steps or explanations.

Here are 5 related questions to deepen your understanding:

1. How can we use the change of base formula to simplify logarithms like $\log_2 320$?
2. Why does the logarithmic property $\log(a \times b) = \log a + \log b$ work, and how is it derived?
3. How can we extend these logarithmic rules to natural logarithms ($\ln$)?
4. What are the real-world applications of logarithms in fields like computer science or physics?
5. How can calculators compute logarithms for non-integer values?

Tip: Remember that simplifying logarithmic expressions often requires recognizing factors that can be written as powers of simpler bases like 2, 3, or 5.