Let's evaluate each logarithmic expression step by step.

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(a) $\log_3(3)$:

• The logarithmic expression $\log_b(x)$ asks the question: "To what power must the base $b$ be raised to get $x$?"
• In this case, $\log_3(3)$ means: "To what power must 3 be raised to get 3?"
• The answer is $1$, since $3^1 = 3$.

So, $\log_3(3) = 1$.

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(b) $\log_9(1)$:

• Any logarithm of 1 is 0, regardless of the base. This is because any non-zero number raised to the power of 0 is 1.
• Therefore, $\log_9(1) = 0$, since $9^0 = 1$.

So, $\log_9(1) = 0$.

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(c) $\log_6(65)$:

• $\log_6(65)$ does not simplify to an exact integer since 65 is not a power of 6.
• To express it in a different way or estimate it, we can use the change of base formula: $\log_b(x) = \frac{\log(x)}{\log(b)}$
• Using this formula: $\log_6(65) = \frac{\log(65)}{\log(6)}$ This would give a decimal approximation, but it's the exact way to express it algebraically.

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Summary:

• $\log_3(3) = 1$
• $\log_9(1) = 0$
• $\log_6(65) = \frac{\log(65)}{\log(6)}$

Would you like me to calculate the decimal approximation for part (c) or give more details on logarithmic properties?

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Related Questions:

1. What is $\log_5(25)$?
2. How do you simplify $\log_7(49)$?
3. What is the general formula for changing the base of a logarithm?
4. How do you solve logarithmic equations like $\log_2(x) = 3$?
5. What is the value of $\log_{10}(1000)$?

Tip:

The logarithm of 1 in any base is always 0, since $b^0 = 1$ for any base $b$.