Pre-calc question: e?

Question

Solve algebraically: e^x + e^-x = 9

Any help would be much appreciated. Please explain so that I can do similar problems on my own. Thanks and 10 points to the best answer.

Answer 1

X e^x eq becomes

e^2x + 1 = 9e^x

let u = e^x, eq becomes

u^2 -9u + 1 = 0

use quad formula to solve for u

then find x

Answer 2

(1) First, multiply both sides by e^x to obtain the following:

e^(2x) + 1 = 9*e^x, a quadratic equation in variable e^x.

(2) subtract 9*e^x from both sides, resulting in the "classic" form of a quadratic equation

e^(2x) - 9*e^x + 1 = 0

(3) Solve this quadratic for the two solutions of e^x:

e^x = [9 + sqrt(81 - 4)]/2 and e^x = [9 - sqrt(81 - 4)]/2

(4) The two results are: e^x = 8.88748 and e^x = 0.112518

(5) Taking the natural log (ln) of these numbers yields

x = + 2.1846

and

x = - 2.1846. [Note: Remember that ln(e^x) = x* ln(e) = x]

Answer 3

this ends up being a quadratic equation...

remember that e^-x = 1/e^x

multiply everything by e^x ==> (e^x)^2 + 1 = 9e^x

but (e^x)^2 = e^(2x)

now you have e^(2x)- 9e^x + 1 = 0

let e^x = u, giving u^2 - 9u + 1 = 0

using the QF:

u = [9 +/- sqrt(81 - 4)] / 2

u = [9 +/- sqrt(77)] / 2

since u = e^x, then x = ln u

so x = ln{[9 +/- sqrt(77)] / 2}

both values inside the ln are positive, so there are two solutions:

x = ln{[9 + sqrt(77)] / 2} and x = ln{[9 - sqrt(77)] / 2}