Stitz-Zeager_College_Algebra_e-book

Hence we get the formula an 1 for our terms and we nd

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Unformatted text preview: seems to be no avoiding substitution and a bit of algebraic unpleasantness. Solving y + 4e2x = 1 for y , we get y = 1 − 4e2x which, when substituted into 2 the second equation, yields 1 − 4e2x + 2ex = 1. After expanding and gathering like terms, we get 16e4x − 8e2x + 2ex = 0. Factoring gives us 2ex 8e3x − 4ex + 1 = 0, and since 2ex = 0 for any real x, we are left with solving 8e3x − 4ex + 1 = 0. We have three terms, and even though this is not a ‘quadratic in disguise’, we can beneﬁt from the substitution u = ex . The equation becomes 8u3 − 4u +1 = 0. Using the techniques set forth in Section 3.3, we ﬁnd u = 1 2 1 is a zero and use synthetic division to factor the left hand side as u − 2 8u2 + 4u − 2 . We use the quadratic formula to solve 8u2 + 4u − 2 = 0 and ﬁnd u = now must solve ex = for ex √ = −1± 5 , 4 1 2 and ex = we ﬁrst note that √ −1± 5 x 4 √ . From e −1− 5 < 0, so ex 4 = = √ −1± 5 . 4 Since u = ex , we 1 1 2 , we get x = ln 2 = − ln(2). √ −1− 5 has no real...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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