can one do this without using calculator 

Yes, you can estimate \(\log_{10}{36}\) without a calculator by using logarithm properties and approximations for known logarithms. Here's how you might do it:

### Step 1: Break Down 36 into Factors
First, express 36 as a product of numbers whose logarithms are easier to calculate:
\[
36 = 6 \times 6 = (2 \times 3) \times (2 \times 3) = 2^2 \times 3^2
\]

### Step 2: Apply Logarithm Properties
Use the logarithm property:
\[
\log_{10}(a \times b) = \log_{10}a + \log_{10}b
\]
This gives:
\[
\log_{10}{36} = \log_{10}(2^2 \times 3^2) = \log_{10}{2^2} + \log_{10}{3^2}
\]

### Step 3: Use Logarithm Powers
Apply the property \(\log_{10}{a^b} = b \log_{10}{a}\):
\[
\log_{10}{36} = 2\log_{10}{2} + 2\log_{10}{3} = 2(\log_{10}{2} + \log_{10}{3})
\]

### Step 4: Approximate Logarithms of 2 and 3
Now, use the approximate values for \(\log_{10}{2}\) and \(\log_{10}{3}\):
\[
\log_{10}{2} \approx 0.3010 \quad \text{and} \quad \log_{10}{3} \approx 0.4771
\]
So:
\[
\log_{10}{36} \approx 2(0.3010 + 0.4771) = 2(0.7781) = 1.5562
\]

### Conclusion
This matches the value we found earlier, \(\log_{10}{36} \approx 1.5563\).

Would you like to explore more logarithm properties or have any further questions?

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Here are 5 related questions for further practice:

1. How can you approximate \(\log_{10}{50}\) using logarithm properties?
2. What is \(\log_{10}{4}\) if you know \(\log_{10}{2}\)?
3. How does the logarithmic scale work for measuring sound (decibels)?
4. Can you express \(\log_{10}{72}\) using logarithms of smaller numbers?
5. What is the inverse operation of finding a logarithm?

**Tip:** Knowing a few key logarithm values, like \(\log_{10}{2}\) and \(\log_{10}{3}\), allows you to estimate many logarithms without a calculator!

Learn how to estimate \( \log_{10}{36} \) without a calculator using logarithm properties and approximations. This problem demonstrates the use of logarithm properties and is suitable for high school students.

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