Let’s solve each logarithmic problem step by step.

Part 1: $\log 394,736$

To estimate:

• $10^5 = 100,000$ and $10^6 = 1,000,000$, so $394,736$ is between $10^5$ and $10^6$. Therefore: $\log 394,736 \text{ is between } 5 \text{ and } 6.$

Approximation: Using a calculator: $\log 394,736 \approx 5.5962.$

Part 2: $\log 995$

To estimate:

• $10^2 = 100$ and $10^3 = 1,000$, so $995$ is between $10^2$ and $10^3$. Therefore: $\log 995 \text{ is between } 2 \text{ and } 3.$

Approximation: Using a calculator: $\log 995 \approx 2.9978.$

Part 3: $\log 0.22$

To estimate:

• $10^{-1} = 0.1$ and $10^0 = 1$, so $0.22$ is between $10^{-1}$ and $10^0$. Therefore: $\log 0.22 \text{ is between } -1 \text{ and } 0.$

Approximation: Using a calculator: $\log 0.22 \approx -0.6576.$

Part 4: $\log 0.0000038$

To estimate:

• $10^{-7} = 0.0000001$ and $10^{-6} = 0.000001$, so $0.0000038$ is between $10^{-7}$ and $10^{-6}$. Therefore: $\log 0.0000038 \text{ is between } -7 \text{ and } -6.$

Approximation: Using a calculator: $\log 0.0000038 \approx -5.4202.$

Let me know if you need additional clarification or would like me to solve the remaining sections.

Here are five related questions for further exploration:

1. How do logarithms relate to exponential functions?
2. Can you calculate $\ln(x)$ instead of $\log(x)$ for similar problems?
3. What are the differences between natural logs and common logs?
4. How does the base of the logarithm affect the result?
5. Can you graph the function $y = \log(x)$ for values of $x > 0$?

Tip: Always estimate the value of a logarithm by bounding it between two powers of 10 before approximating.