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Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 7, Problem 7.1.68

Logarithm properties Use the integral definition of the natural logarithm to prove that ln(x/y) = ln x - ln y.

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Recall the integral definition of the natural logarithm: for any positive number \( a \), \( \ln a = \int_1^a \frac{1}{t} \, dt \).
Express \( \ln \left( \frac{x}{y} \right) \) using the integral definition: \( \ln \left( \frac{x}{y} \right) = \int_1^{\frac{x}{y}} \frac{1}{t} \, dt \).
Use a substitution to rewrite the integral with limits from 1 to \( x \) and 1 to \( y \). Let \( t = \frac{u}{y} \), so that when \( t = 1 \), \( u = y \), and when \( t = \frac{x}{y} \), \( u = x \).
Rewrite the integral in terms of \( u \): \( \int_1^{\frac{x}{y}} \frac{1}{t} \, dt = \int_y^x \frac{1}{\frac{u}{y}} \cdot \frac{1}{y} \, du = \int_y^x \frac{y}{u} \cdot \frac{1}{y} \, du = \int_y^x \frac{1}{u} \, du \).
Split the integral \( \int_y^x \frac{1}{u} \, du \) into \( \int_1^x \frac{1}{u} \, du - \int_1^y \frac{1}{u} \, du \), which equals \( \ln x - \ln y \), thus proving \( \ln \left( \frac{x}{y} \right) = \ln x - \ln y \).

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