Textbook Question
Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume x > 0 and y > 0.
b. ln 0 = 1
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0. Functions
Logarithmic Functions
1:42 minutesProblem 7.1.74c
Textbook Question
ln x is unbounded Use the following argument to show that lim (x → ∞) ln x = ∞ and lim (x → 0⁺) ln x = −∞.
c. Show that ln 2ⁿ > n/2 and ln 2^(−n) < −n/2.
Verified step by step guidance1
Recall the properties of logarithms, specifically that \( \ln(a^b) = b \ln a \). Use this to rewrite \( \ln 2^n \) as \( n \ln 2 \) and \( \ln 2^{-n} \) as \( -n \ln 2 \).
Note that \( \ln 2 \) is a positive constant since 2 is greater than 1. To show \( \ln 2^n > \frac{n}{2} \), compare \( n \ln 2 \) with \( \frac{n}{2} \). This reduces to showing \( \ln 2 > \frac{1}{2} \).
Similarly, to show \( \ln 2^{-n} < -\frac{n}{2} \), compare \( -n \ln 2 \) with \( -\frac{n}{2} \). This reduces to showing \( \ln 2 > \frac{1}{2} \) again, which is consistent with the previous step.
Since \( \ln 2 \) is approximately 0.693, which is greater than 0.5, both inequalities hold true for all positive integers \( n \).
These inequalities help illustrate the behavior of \( \ln x \) as \( x \to \infty \) and \( x \to 0^+ \) by bounding the logarithm with linear functions that tend to infinity or negative infinity, respectively.
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Video duration:
1mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Behavior of the Natural Logarithm Function
The natural logarithm function, ln(x), increases without bound as x approaches infinity, and decreases without bound as x approaches zero from the right. This means lim (x → ∞) ln(x) = ∞ and lim (x → 0⁺) ln(x) = −∞, reflecting its unbounded growth and decay.
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Derivative of the Natural Logarithmic Function
Properties of Exponents and Logarithms
Logarithms and exponents are inverse operations. For any positive base a, ln(a^n) = n ln(a). This property allows us to rewrite expressions like ln(2^n) as n ln(2), facilitating comparisons and inequalities involving exponential and logarithmic terms.
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Inequalities Involving Logarithms
To show inequalities such as ln(2^n) > n/2, one uses known bounds or approximations of ln(2). Since ln(2) ≈ 0.693 > 0.5, multiplying by n preserves the inequality. Similarly, for negative exponents, ln(2^{-n}) = -n ln(2) < -n/2, demonstrating how logarithmic inequalities relate to linear bounds.
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