https://file.aiquickdraw.com/mathbot/file/1740456400593v7ny0c0k.jpg

Let’s work through each of these logarithmic expressions step by step:

1. **log₃(1/27)**  
   Since 27 = 3³, we can express 1/27 as 3⁻³.  
   \[
   \log_3(3^{-3}) = -3
   \]
   So, log₃(1/27) = -3.

2. **log₆(√6)**  
   We can rewrite the square root of 6 as 6^(1/2).  
   Using the logarithmic rule logₐ(b^x) = x * logₐ(b), we get:
   \[
   \log_6(6^{1/2}) = \frac{1}{2}
   \]
   So, log₆(√6) = 1/2.

3. **log₂(1/8) +**  
   For this expression, we know that 8 = 2³, so 1/8 = 2⁻³.  
   \[
   \log_2(2^{-3}) = -3
   \]
   So, log₂(1/8) = -3. The "+" symbol here seems to imply that more information follows, but it’s unclear from the image.

Would you like me to continue solving the problem or explain in more detail? Feel free to ask!

**Related questions:**
1. How can logarithmic rules help simplify expressions like these?
2. What happens when the base of a logarithm is the same as the argument?
3. Can we apply the same techniques to solve logarithms with different bases?
4. What’s the difference between logarithms with integer bases and irrational bases?
5. How would you solve a logarithmic equation involving multiple terms?

**Tip**: When working with logarithms, it's often helpful to express the argument in terms of the base to simplify the expression!

log₃(1/27), log₆(√6), log₂(1/8) +

Learn how to solve logarithmic expressions like log₃(1/27), log₆(√6), and log₂(1/8) with step-by-step explanations. This tutorial covers logarithmic properties, exponentiation, and change of base formula. Suitable for students in Grades 9-11.

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