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QUOTIENT PROPERTY log, (u)-log, (v) = log V

QUOTIENT PROPERTY log, (u)-log, (v) = log V

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Please complete where it says "quotient property"

Calculus Connections Project 1: Proving Properties of Logarithmic Functions (Using Properties of Exponential Functions) In previous classes, we have stated that logarithmic and exponential functions are inverses of each other. We have noticed especially that rules were previously established for exponential functions have a logarithmic counterpart. For example, and exponential function that has not undergone any transformations has a y-intercept at (0, 1). Similarly, a logarithmic function that has not undergone any transformations has an x-intercept at (1, 0). In addition to the ideas seen using "switch and solve," other properties have counterparts. Here we will look at the product property of logarithms: PRODUCT PROPERTY log (u) +log, (v) = log (uv) Let's begin with log(u)+log (v) = y, where y is some unknown quantity. Let us now individually solve each logarithm. This yields the following: log(u)=y-log, (v) log, (v)=y-log, (u) Now let us change each logarithm into an exponential. This will give: a."-log, (v) = 11 a."-log, (1) = V ¹-log, (v)-log(u) a Name: and Remember that ultimately, we want to get log, (uv). So keeping this in mind, let us multiply these two exponential together. Doing this, we get: = UV -log, (v) [y-log (1) and Thinking back to the properties of exponential functions, though, we should be able to recall that when two like bases are multiplied together, the exponents are added. Here, we have like bases, a, on the left-hand side of the equation. Thus, we will add the exponents together. This will give the following: =uva j²y—(log, (v) + log„ (4)) =UIV
Transcribed Image Text:Calculus Connections Project 1: Proving Properties of Logarithmic Functions (Using Properties of Exponential Functions) In previous classes, we have stated that logarithmic and exponential functions are inverses of each other. We have noticed especially that rules were previously established for exponential functions have a logarithmic counterpart. For example, and exponential function that has not undergone any transformations has a y-intercept at (0, 1). Similarly, a logarithmic function that has not undergone any transformations has an x-intercept at (1, 0). In addition to the ideas seen using "switch and solve," other properties have counterparts. Here we will look at the product property of logarithms: PRODUCT PROPERTY log (u) +log, (v) = log (uv) Let's begin with log(u)+log (v) = y, where y is some unknown quantity. Let us now individually solve each logarithm. This yields the following: log(u)=y-log, (v) log, (v)=y-log, (u) Now let us change each logarithm into an exponential. This will give: a."-log, (v) = 11 a."-log, (1) = V ¹-log, (v)-log(u) a Name: and Remember that ultimately, we want to get log, (uv). So keeping this in mind, let us multiply these two exponential together. Doing this, we get: = UV -log, (v) [y-log (1) and Thinking back to the properties of exponential functions, though, we should be able to recall that when two like bases are multiplied together, the exponents are added. Here, we have like bases, a, on the left-hand side of the equation. Thus, we will add the exponents together. This will give the following: =uva j²y—(log, (v) + log„ (4)) =UIV
However, recall from before that log (u) +log (v) = y. Substituting this into our new equation, we get: a²y Now we will change this exponential into a logarithm. Doing so gives the following: log (uv) = y =uv⇒ a' = uv Therefore, since log (uv) = y and log (u)+log (v) = y, we get log, (u) +log (v) = log (uv), which is the product property! Now it's your turn. Keeping in mind the steps and procedures from the product property example, show why the quotient property and power property are true. QUOTIENT PROPERTY log, (u)-log, (v) = log₁| 21 V
Transcribed Image Text:However, recall from before that log (u) +log (v) = y. Substituting this into our new equation, we get: a²y Now we will change this exponential into a logarithm. Doing so gives the following: log (uv) = y =uv⇒ a' = uv Therefore, since log (uv) = y and log (u)+log (v) = y, we get log, (u) +log (v) = log (uv), which is the product property! Now it's your turn. Keeping in mind the steps and procedures from the product property example, show why the quotient property and power property are true. QUOTIENT PROPERTY log, (u)-log, (v) = log₁| 21 V
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