Stitz-Zeager_College_Algebra_e-book

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online