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2. Solving the following: a. log5 (2x + 1) = 1 − log5 (x + 2) b. 32x = 72x+1

2. Solving the following: a. log5 (2x + 1) = 1 − log5 (x + 2) b. 32x = 72x+1

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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2. Solving the following: a. log5 (2x + 1) = 1 − log5 (x + 2) b. 3²x = 72x+1
Transcribed Image Text:2. Solving the following: a. log5 (2x + 1) = 1 − log5 (x + 2) b. 3²x = 72x+1
1. The displacement, s, in metres, of a particle moving back and forth in a straight line can be modelled by the function s(t) = 4t² − 10t + 13 where t is measured in seconds. a) Find the average rate of change of the distance with respect to time from 2 s to 5 s. b) Determine the approximate value for the instantaneous rate of change of the displacement of the particle at 2 s.
Transcribed Image Text:1. The displacement, s, in metres, of a particle moving back and forth in a straight line can be modelled by the function s(t) = 4t² − 10t + 13 where t is measured in seconds. a) Find the average rate of change of the distance with respect to time from 2 s to 5 s. b) Determine the approximate value for the instantaneous rate of change of the displacement of the particle at 2 s.
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