Math Problem Statement

number 30

log_3(1/9) - 3log_3(3)

Solution

Let's solve problem 30 from the image step by step.

Problem:

log3(19)3log3(3)\log_3\left(\frac{1}{9}\right) - 3\log_3(3)

Step 1: Apply logarithmic properties.

  1. Simplify log3(19)\log_3\left(\frac{1}{9}\right).

    We know that 19=32\frac{1}{9} = 3^{-2}. Using the logarithmic property logb(an)=nlogb(a)\log_b(a^n) = n \log_b(a), we get: log3(19)=log3(32)=2log3(3)\log_3\left(\frac{1}{9}\right) = \log_3(3^{-2}) = -2 \log_3(3) Since log3(3)=1\log_3(3) = 1, this simplifies to: 2log3(3)=21=2-2 \log_3(3) = -2 \cdot 1 = -2

  2. Simplify 3log3(3)3\log_3(3).

    From the logarithmic property nlogb(a)=logb(an)n \log_b(a) = \log_b(a^n), and knowing that log3(3)=1\log_3(3) = 1, we get: 3log3(3)=31=33 \log_3(3) = 3 \cdot 1 = 3

Step 2: Combine the results.

Now, we can subtract the two terms: 23=5-2 - 3 = -5

Final Answer:

log3(19)3log3(3)=5\log_3\left(\frac{1}{9}\right) - 3\log_3(3) = -5

Would you like more details on any step, or do you have any questions?

Here are 5 related questions to deepen your understanding:

  1. How would you evaluate log3(279)\log_3\left(\frac{27}{9}\right) using properties of logarithms?
  2. Can you simplify logb(1bn)\log_b\left(\frac{1}{b^n}\right) in general terms?
  3. What happens when you evaluate logb(1)\log_b(1)? Why is it always 0?
  4. How can you use the change-of-base formula to evaluate logarithms with bases other than 10 or ee?
  5. How would you simplify nlogb(b)n \log_b(b) for any nn?

Tip:

Whenever you see a logarithm of a fraction, try expressing the fraction in terms of powers of the base, as this often simplifies the problem quickly.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Exponents

Formulas

log_b(a^n) = n log_b(a)
log_b(b) = 1

Theorems

Logarithmic properties
Exponent rules

Suitable Grade Level

Grades 10-12