**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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