Math Problem Statement

log base 6 2 + log base 6 18 =

Solution

To simplify this expression:

log62+log618\log_6 2 + \log_6 18

we can use the property of logarithms that states:

logbA+logbB=logb(A×B)\log_b A + \log_b B = \log_b (A \times B)

So,

log62+log618=log6(2×18)\log_6 2 + \log_6 18 = \log_6 (2 \times 18)

Calculating inside the logarithm:

2×18=362 \times 18 = 36

Thus,

log62+log618=log636\log_6 2 + \log_6 18 = \log_6 36

Since 3636 is 626^2, we can rewrite it as:

log636=log6(62)=2\log_6 36 = \log_6 (6^2) = 2

Final Answer:

log62+log618=2\log_6 2 + \log_6 18 = 2

Would you like a further breakdown or have any questions?


Here are five related questions to deepen your understanding:

  1. How do you simplify logbAlogbB\log_b A - \log_b B?
  2. What is log612+log63\log_6 12 + \log_6 3?
  3. How would you convert log62\log_6 2 to a different base, like base 10?
  4. What is log66\log_6 6 and why?
  5. How do you evaluate logb(Ak)\log_b (A^k) in general?

Tip: When adding logs with the same base, remember you can combine them by multiplying the values inside the logarithm.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Logarithmic Properties
Exponentiation

Formulas

log_b A + log_b B = log_b (A * B)
log_b (A^k) = k * log_b A

Theorems

Product Property of Logarithms

Suitable Grade Level

Grades 9-12