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We are given that \( \log_{10} 4 = 0.6021 \) and asked to find \( \log_{10} \sqrt{400} \) without using a calculator. Let's break down the problem step by step.

### Step 1: Simplify \( \log_{10} \sqrt{400} \)
We can rewrite the square root using exponents:

\[
\sqrt{400} = 400^{1/2}
\]

Thus:

\[
\log_{10} \sqrt{400} = \log_{10} (400^{1/2}) = \frac{1}{2} \log_{10} 400
\]

### Step 2: Simplify \( \log_{10} 400 \)
We know that:

\[
400 = 4 \times 100
\]

So we can use the logarithmic property \( \log(ab) = \log a + \log b \) to get:

\[
\log_{10} 400 = \log_{10} 4 + \log_{10} 100
\]

We already know that \( \log_{10} 4 = 0.6021 \), and since \( \log_{10} 100 = 2 \) (because \( 100 = 10^2 \)), we can substitute:

\[
\log_{10} 400 = 0.6021 + 2 = 2.6021
\]

### Step 3: Compute \( \frac{1}{2} \log_{10} 400 \)
Now, we need to divide the result by 2:

\[
\log_{10} \sqrt{400} = \frac{1}{2} \times 2.6021 = 1.30105
\]

### Final Answer:
\[
\log_{10} \sqrt{400} \approx 1.3011
\]
(Rounding to four decimal places.)

Would you like more details or have any further questions?

Here are 5 related questions to explore this topic:
1. How would you compute \( \log_{10} \sqrt{625} \) without a calculator?
2. What is the logarithmic property that allows us to simplify \( \log_{10} \sqrt{400} \)?
3. How can you express logarithms with base \( e \) (natural logarithms)?
4. Can you use the same approach to solve \( \log_{10} \sqrt{50} \)?
5. How would this problem change if the logarithm was base 2 instead of base 10?

**Tip:** Remember that logarithmic properties, such as \( \log(a \times b) = \log a + \log b \) and \( \log(a^b) = b \log a \), are essential tools in simplifying expressions.

Given log10 4 = 0.6021, find the logarithm without using a calculator. log10 √400 = ? (Round to four decimal places as needed.)

Learn how to calculate log10 √400 given log10 4 = 0.6021, using properties of logarithms and exponents, and solve without a calculator. Step-by-step instructions with key formulas and rounding to four decimal places. Suitable for high school students in Grades 9-12.

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