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Stitz-Zeager_College_Algebra_e-book

# While we shall not have much need of these properties

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Unformatted text preview: solutions. We 4 As are 536 Systems of Equations and Matrices √ √ left with ex = −1+ 5 , so that x = ln −1+ 5 . We now return to y = 1 − 4e2x to ﬁnd the 4 4 accompanying y values for each of our solutions for x. For x = − ln(2), we get y= = = = = For x = ln √ −1+ 5 4 1 − 4e2x 1 − 4e−2 ln(2) 1 − 4eln(1/4) 1−4 1 4 0 , we have y = 1 − 4e2x 2 ln = 1 − 4e ln = 1 − 4e = 1−4 = 1−4 = √ −1+ 5 2 √ √ −1+ 5 4 √ −1+ 5 4 2 √2 −1+ 5 4 √ 3− 5 8 √ We get two solutions, (0, − ln(2)), ln −1+ 5 , −1+ 5 . It is a good review of the prop4 2 erties of logarithms to verify both solutions, so we leave that to the reader. We are able to sketch y = 1 − 4e2x using transformations, but the second equation is more diﬃcult and we resort to the calculator. We note that to graph y 2 + 2ex = 1, we need to graph both the √ positive and negative roots, y = ± 1 − 2ex . After some careful zooming,2 we conﬁrm our solutions. √ The graphs of y = 1 − 4e2x and y = ± 1 − 2ex . 3. Our last system involves three variables and gives some insight on how to keep such systems organized...
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