Let me transcribe and solve the logarithmic expressions from the image.

Transcribed Problems:

1. $\log_2 16 + \log_2 100$
2. $\log_5 98 - \log_5 4$
3. $2\log_3 9 + \log_3 81$

Solution:

1. $\log_2 16 + \log_2 100$

Using the logarithmic property $\log_b(m) + \log_b(n) = \log_b(m \cdot n)$: $\log_2 16 + \log_2 100 = \log_2 (16 \cdot 100) = \log_2 1600$ Next, express $1600$ as a power of $2$: $1600 = 2^6 \cdot 25 = 2^6 \cdot 5^2$ Thus: $\log_2 1600 = 6 + \log_2(25) = 6 + \frac{\log(25)}{\log(2)} \quad \text{(if exact value required)}.$

2. $\log_5 98 - \log_5 4$

Using the logarithmic property $\log_b(m) - \log_b(n) = \log_b\left(\frac{m}{n}\right)$: $\log_5 98 - \log_5 4 = \log_5\left(\frac{98}{4}\right)$ Simplify: $\frac{98}{4} = 24.5 \quad \text{so} \quad \log_5\left(\frac{98}{4}\right) = \log_5(24.5)$

3. $2\log_3 9 + \log_3 81$

Use the property $a\log_b(m) = \log_b(m^a)$: $2\log_3 9 = \log_3(9^2) = \log_3(81)$ So: $2\log_3 9 + \log_3 81 = \log_3(81) + \log_3(81)$ Using the addition property: $\log_3(81) + \log_3(81) = \log_3(81 \cdot 81) = \log_3(6561)$ Since $6561 = 3^8$: $\log_3(6561) = 8$

Final Answers:

1. $\log_2 1600 = 6 + \log_2(25)$ or $6 + \frac{\log(25)}{\log(2)}$.
2. $\log_5(24.5)$.
3. $8$.

Would you like detailed steps for any part of these solutions or approximations for logarithms?

Related Questions:

1. How are logarithmic properties applied to simplify expressions?
2. What are common techniques to estimate logarithmic values?
3. Can you generalize the use of $\log_b(m^n) = n\log_b(m)$?
4. How does change of base simplify solving logs like $\log_5(24.5)$?
5. Why is $\log_b(1) = 0$ always true?

Tip:

Always simplify logarithmic arguments to their prime factors for quicker calculations.