Chapter 6: Problem 74
Use the Change-of-Base Formula and a calculator to evaluate each logarithm. Round your answer to three decimal places. \(\log _{1 / 2} 15\)
Short Answer
Expert verified
-3.907
Step by step solution
01
Identify the Change-of-Base Formula
The Change-of-Base Formula is: \[ \log_b{a} = \frac{ \log_c{a} }{ \log_c{b} } \] We can choose any base (often base 10 or base e) to convert the logarithm.
02
Apply the Change-of-Base Formula
To evaluate \log_{1/2}{15}\ using the change-of-base formula with base 10, rewrite as follows: \[ \log_{1/2}{15} = \frac{ \log_{10}{15} }{ \log_{10}{1/2} } \]
03
Calculate the Numerator
Use a calculator to find the logarithm of 15 with base 10: \[ \log_{10}{15} \approx 1.176 \]
04
Calculate the Denominator
Use a calculator to find the logarithm of 1/2 with base 10: \[ \log_{10}{1/2} \approx -0.301 \]
05
Divide the Values
Now divide the results obtained: \[ \log_{1/2}{15} = \frac{1.176}{-0.301} \approx -3.907 \]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Logarithms
Logarithms are mathematical tools that help us determine the power to which a number (base) must be raised to obtain another number. They are the inverse operations of exponentiation. For example, if we have \(2^3 = 8\), then the logarithm base 2 of 8 is written as \( \text{log}_2 8 \) and equals 3.
They allow us to simplify complex calculations, especially those involving multiplication, division, and exponentiation. They are widely used in various fields such as science, engineering, and financial modeling.
Logarithms come with properties that make them quite useful, like:
They allow us to simplify complex calculations, especially those involving multiplication, division, and exponentiation. They are widely used in various fields such as science, engineering, and financial modeling.
Logarithms come with properties that make them quite useful, like:
- Product Rule: \( \text{log}_b(xy) = \text{log}_b(x) + \text{log}_b(y) \)
- Quotient Rule: \( \text{log}_b(x/y) = \text{log}_b(x) - \text{log}_b(y) \)
- Power Rule: \( \text{log}_b(x^r) = r \text{log}_b(x) \)
Base Conversion
Base conversion is a useful technique for evaluating logarithms when the given base is not common, like 10 (common logarithm) or e (natural logarithm). This is where the Change-of-Base Formula comes into play. The formula is expressed as:
\( \text{log}_b{a} = \frac{ \text{log}_c{a} }{ \text{log}_c{b} } \)
Here, \( b \) and \( c \) are the bases, and \( a \) is the argument we want the logarithm of. Often, we select base 10 or base e because these are easy to compute using calculators.
For example, to convert \( \text{log}_{1/2}{15} \) to base 10, the formula alters the expression to:
\( \text{log}_b{a} = \frac{ \text{log}_c{a} }{ \text{log}_c{b} } \)
Here, \( b \) and \( c \) are the bases, and \( a \) is the argument we want the logarithm of. Often, we select base 10 or base e because these are easy to compute using calculators.
For example, to convert \( \text{log}_{1/2}{15} \) to base 10, the formula alters the expression to:
- \( \text{log}_{1/2}{15} = \frac{ \text{log}_{10}{15} }{ \text{log}_{10}{1/2} } \)
Calculator Use
Using a calculator is essential when dealing with logarithms, especially for non-standard bases that require conversion. Most scientific calculators have dedicated logarithm functions for base 10 (log) and base e (ln).
To evaluate \( \text{log}_{10}{15} \), simply input 15 and press the 'log' button:
\( \frac{1.176}{-0.301} \) which approximately equals -3.907.
Ensure accuracy by double-checking these values, and remember to round your answers to the desired decimal places as instructed by the problem. Calculators make it significantly easier to work with complex expressions and non-standard logarithmic bases.
To evaluate \( \text{log}_{10}{15} \), simply input 15 and press the 'log' button:
- \( \text{log}_{10}{15} \) will yield approximately 1.176
- \( \text{log}_{10}{1/2} \) will yield approximately -0.301
\( \frac{1.176}{-0.301} \) which approximately equals -3.907.
Ensure accuracy by double-checking these values, and remember to round your answers to the desired decimal places as instructed by the problem. Calculators make it significantly easier to work with complex expressions and non-standard logarithmic bases.