Problem 102
Question
Use the change-of-base formula to find logarithm to four decimal places. \(\log _{\pi} e\)
Step-by-Step Solution
Verified Answer
\(\log_\pi e \approx 0.8731\)
1Step 1: Understand the Change-of-Base Formula
The change-of-base formula allows you to evaluate a logarithm with any base by expressing it in terms of logarithms with a different base, usually base 10 or base e (the natural logarithm). The formula is: \[\log_b a = \frac{\log_k a}{\log_k b}\]where \(b\) is the base of the original logarithm, \(a\) is the value whose logarithm you want to find, and \(k\) is the new base you choose (commonly 10 or \(e\)).
2Step 2: Apply the Change-of-Base Formula
In this exercise, we want to find \(\log_\pi e\). Using the change-of-base formula, we choose base \(e\) (natural logarithm) for convenience:\[\log_\pi e = \frac{\ln e}{\ln \pi}\]We apply the natural logarithm \(\ln\) to both \(e\) and \(\pi\).
3Step 3: Evaluate the Natural Logarithms
Calculate both \(\ln e\) and \(\ln \pi\): - \(\ln e = 1\), because the natural logarithm of Euler's number \(e\) is 1 by definition.- Use a calculator to find \(\ln \pi\), which is approximately 1.1447 to four decimal places.
4Step 4: Calculate \(\log_\pi e\)
Substitute the evaluated logarithm values into the change-of-base formula: \[\log_\pi e = \frac{1}{1.1447}\]Calculate the result to find \(\log_\pi e \approx 0.8731\) when rounded to four decimal places.
Key Concepts
LogarithmsNatural LogarithmBase Conversion
Logarithms
Logarithms are mathematical tools used to solve for powers. They are the inverse operation of exponentiation, meaning that they help you find the exponent that a base number is raised to in order to reach another number. For example, if you know that \(b^c = a\) (where \(b\) is the base, \(c\) is the exponent, and \(a\) is the result), then \(\log_b a = c\) tells you what \(c\) is.
- Base: The number that is raised to a power.
- Exponent: The power to which the base number is raised.
- Logarithm: The exponent itself.
Natural Logarithm
Natural logarithms are a specific kind of logarithm with a base of the special number \(e\). The number \(e\) is an irrational and transcendental number approximately equal to 2.71828.
They are denoted as \(\ln\) and are particularly compelling for calculations involving continuous growth processes like compound interest or population growth.
They are denoted as \(\ln\) and are particularly compelling for calculations involving continuous growth processes like compound interest or population growth.
- Base \(e\): A constant that appears widely in mathematics, especially in calculus.
- Symbol \(\ln\): Used to denote natural logarithms.
Base Conversion
Base conversion in terms of logarithms involves changing from one logarithmic base to another, often for ease of calculation. This is particularly useful because calculators typically feature base 10 (common logarithms) and natural logarithms (base \(e\)), but not for other bases like \(\pi\).
The change-of-base formula is:
\[\log_b a = \frac{\log_k a}{\log_k b}\]Here, \(b\) is the old base, \(a\) is the number you're finding the logarithm of, and \(k\) is the new base. This formula permits the calculation of any logarithm using your desired base (commonly 10 or \(e\)).
The change-of-base formula is:
\[\log_b a = \frac{\log_k a}{\log_k b}\]Here, \(b\) is the old base, \(a\) is the number you're finding the logarithm of, and \(k\) is the new base. This formula permits the calculation of any logarithm using your desired base (commonly 10 or \(e\)).
- Old Base (\(b\)): The original base of your logarithm.
- Value (\(a\)): The number you need the logarithm for.
- New Base (\(k\)): The base to which you're converting.
Other exercises in this chapter
Problem 101
Use the change-of-base formula to find logarithm to four decimal places. \(\log _{\sqrt{2}} \sqrt{5}\)
View solution Problem 101
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View solution Problem 102
After participating in an eight-week weight training program, a college student was selected to be part of a study to see how much strength he would lose if he
View solution Problem 103
Tritium Decay. The half-life of tritium is 12.4 years. How long will it take for \(25 \%\) of a sample of tritium to decompose?
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