Math Problem Statement

Evaluate. log Subscript 3 Baseline StartFraction 1 Over 243 EndFraction Question content area bottom Part 1 log Subscript 3 Baseline StartFraction 1 Over 243 EndFraction equals    enter your response here

Solution

To evaluate log3(1243)\log_3 \left( \frac{1}{243} \right), let's proceed step by step:

Step 1: Express 1243\frac{1}{243} as a power of 3.

We know that 243=35243 = 3^5. Therefore: 1243=35\frac{1}{243} = 3^{-5}

Step 2: Use the logarithmic property.

We apply the logarithmic rule logb(ac)=clogb(a)\log_b (a^c) = c \cdot \log_b (a). In this case: log3(35)=5log3(3)\log_3 \left( 3^{-5} \right) = -5 \cdot \log_3 (3)

Step 3: Simplify using log3(3)=1\log_3 (3) = 1.

Thus: 51=5-5 \cdot 1 = -5

Final answer:

log3(1243)=5\log_3 \left( \frac{1}{243} \right) = -5

Would you like further details or explanations?

Here are 5 related questions:

  1. What is log3(243)\log_3 (243)?
  2. How do you simplify logb(1/a)\log_b (1/a) for any base bb?
  3. What is log2(132)\log_2 \left( \frac{1}{32} \right)?
  4. Can you explain the change of base formula for logarithms?
  5. How do logarithms behave when the base is a fraction?

Tip: When evaluating logarithms of fractions, express the fraction in terms of a power of the base to simplify.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Exponents
Algebra

Formulas

log_b (a^c) = c * log_b (a)
log_b (b) = 1

Theorems

Logarithmic Properties

Suitable Grade Level

Grades 9-11