Q1. log 3600 is equal to

a. 2log6 +1

b. 6log2 + 1

c. 2log6 +2

d. 6log2 + 2

To solve \( \log 3600 \), we begin by breaking down the number 3600 into its prime factorization:

\[
3600 = 36 \times 100 = (6^2) \times (10^2)
\]
Now apply the logarithm:

\[
\log 3600 = \log [(6^2) \times (10^2)]
\]

Using the logarithmic property \( \log(ab) = \log a + \log b \) and \( \log(a^n) = n \log a \):

\[
\log 3600 = \log (6^2) + \log (10^2)
\]
\[
\log 3600 = 2 \log 6 + 2 \log 10
\]
Since \( \log 10 = 1 \), this simplifies to:

\[
\log 3600 = 2 \log 6 + 2
\]

Thus, the correct answer is:

\[
\boxed{c. \ 2\log6 + 2}
\]

Would you like further details or have any questions?

Here are 5 related questions:
1. How do you simplify expressions involving logarithmic properties?
2. What is the logarithmic property of powers, and how does it apply to log equations?
3. How can you convert logarithms with different bases?
4. Why does \( \log 10 = 1 \) in base-10 logarithms?
5. How would you evaluate \( \log 3600 \) using natural logarithms?

**Tip:** Always remember logarithmic properties like \( \log(ab) = \log a + \log b \) and \( \log(a^n) = n \log a \) to simplify complex expressions easily.

Q1. log 3600 is equal to
a. 2log6 + 1
b. 6log2 + 1
c. 2log6 + 2
d. 6log2 + 2

Learn how to solve the logarithmic equation log 3600 using prime factorization and logarithmic properties. This problem focuses on simplifying logarithmic expressions, making use of key concepts such as log(ab) = log a + log b and log(a^n) = n log a. Suitable for students in Grades 9-11.

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