Table of contents
- 0. Review of Algebra4h 16m
- 1. Equations & Inequalities3h 18m
- 2. Graphs of Equations43m
- 3. Functions2h 17m
- 4. Polynomial Functions1h 44m
- 5. Rational Functions1h 23m
- 6. Exponential & Logarithmic Functions2h 28m
- 7. Systems of Equations & Matrices4h 6m
- 8. Conic Sections2h 23m
- 9. Sequences, Series, & Induction1h 19m
- 10. Combinatorics & Probability1h 45m
6. Exponential & Logarithmic Functions
Introduction to Logarithms
Problem 143
Textbook Question
Without using a calculator, find the exact value of: [log3 81 - log𝝅 1]/[log2√2 8 - log 0.001]
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1
Step 1: Recognize that \( \log_3 81 \) can be simplified using the fact that 81 is a power of 3. Specifically, \( 81 = 3^4 \), so \( \log_3 81 = 4 \).
Step 2: Understand that \( \log_\pi 1 \) is 0 because any logarithm of 1 is 0, regardless of the base.
Step 3: Simplify the expression in the numerator: \( \log_3 81 - \log_\pi 1 = 4 - 0 = 4 \).
Step 4: For the denominator, recognize that \( \log_{\sqrt{2}} 8 \) can be rewritten using the change of base formula: \( \log_{\sqrt{2}} 8 = \frac{\log_2 8}{\log_2 \sqrt{2}} \). Since \( 8 = 2^3 \), \( \log_2 8 = 3 \). Also, \( \sqrt{2} = 2^{1/2} \), so \( \log_2 \sqrt{2} = \frac{1}{2} \). Therefore, \( \log_{\sqrt{2}} 8 = \frac{3}{1/2} = 6 \).
Step 5: Recognize that \( \log 0.001 \) is \( \log 10^{-3} = -3 \). Therefore, the denominator becomes \( 6 - (-3) = 6 + 3 = 9 \). Finally, the expression simplifies to \( \frac{4}{9} \).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Logarithmic Properties
Logarithmic properties are rules that simplify the manipulation of logarithms. Key properties include the product rule (log_b(mn) = log_b(m) + log_b(n)), the quotient rule (log_b(m/n) = log_b(m) - log_b(n)), and the power rule (log_b(m^k) = k * log_b(m)). Understanding these properties is essential for simplifying expressions involving logarithms.
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Change of Base Property
Change of Base Formula
The change of base formula allows the conversion of logarithms from one base to another, expressed as log_b(a) = log_k(a) / log_k(b) for any positive k. This is particularly useful when dealing with logarithms of bases that are not easily computable, enabling the evaluation of logarithmic expressions in a more manageable form.
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Change of Base Property
Exponential and Logarithmic Relationships
Exponential and logarithmic functions are inverses of each other. For example, if b^y = x, then log_b(x) = y. This relationship is crucial for solving logarithmic equations and understanding how to manipulate logarithmic expressions, as it provides a foundation for converting between exponential and logarithmic forms.
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Solving Logarithmic Equations
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