Advanced Placement Calculus Graphical Numerical Algebraic Sixth Edition High School Binding Copyright 2020
6th Edition
ISBN: 9781418300203
Author: Prentice Hall
Publisher: Prentice Hall
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Question
Chapter 0.5, Problem 33E
To determine
To solve: An equation algebrically and verify solution graphically.
Expert Solution & Answer
Answer to Problem 33E
The required algebraic solution is t=15.744 , and graphical t=15.747 .
Explanation of Solution
Given data: The given equation is (1.045)t=2 .
Formula used:
The log of a number am to the base 10 is log(am)=mlog(a) .
Calculation:
The given equation is (1.045)t=2 .Taking log of both sides,
log(1.045)t=log2t⋅log(1.045)=log2} … (1)
Since, log(1.045)=0.01912 and log2=0.30103 , so substitute 0.01912 for log(1.045) and 0.30103 for log2 in the result (1),
t⋅(0.01912)=0.30103t=0.301030.01912t=15.744
Write given equation in functional form as,
y(t)=(1.045)t−2
The solution y(t)=0 is the solution of the equation (1.045)t−2=0 or (1.045)t=2 .
The graph of the function y(t)=(1.045)t−2 is shown in Fig. 1.
The curve y(t)=(1.045)t−2 intersect t-axis at point t=15.747 , y(t)=0 .
Hence, t=15.747 is the solution of the equation (1.045)t=2 .
Thus, the algebraic solution of the equation (1.045)t=2 is t=15.744 , and the graphical solution t=15.747 .
Therefore, the solution of the equation (1.045)t=2 is t=15.74 ,
Chapter 0 Solutions
Advanced Placement Calculus Graphical Numerical Algebraic Sixth Edition High School Binding Copyright 2020
Ch. 0.1 - Prob. 1QRCh. 0.1 - Prob. 2QRCh. 0.1 - Prob. 3QRCh. 0.1 - Prob. 4QRCh. 0.1 - Prob. 5QRCh. 0.1 - Prob. 6QRCh. 0.1 - Prob. 7QRCh. 0.1 - Prob. 8QRCh. 0.1 - Prob. 9QRCh. 0.1 - Prob. 10QR
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