Help on math?

Question

ok... e is euler's number

(e^x + e^-x)/2 = 3

I'm not sure how to solve this... because if you added em wouldn't it be e^0 = 6... but 1 doesn't equal three... and then i can't find x

there's a hint in my book that says

[Multiply each side by e^x.]... which just confuses me more.

Answer 1

(e^x + e^-x) / 2 = 3

e^x + e^-x = 6

e^-x(1 + e^(2x)) = 6 Factor out e^-x

e^x [e^-x (1 + e^(2x))] = (e^x) (6) Multiply both sides by e^x to factor out e^-x ( any variable to the -x power mutiplied by itself to the x power will give you the variable to the zero power which is one.)

1+ e^(2x) = 6e^x simplify

1 + e^(2x) - 6e^x = 0 subtract 6e^x

1 + e^x (e^x - 6) = 0 Factor out e^x

Let e^x = n

1 + n (n - 6) = 0

n^2 - 6n + 1 = 0 simplify

n = {[6 +or- ((-6)^2 - 4(1)(1))^(1/2)] / 2(1) Quadratic formula

n ={6 +or- 32^(1/2)} / 2 Simplify
n = {6 +or- 4(2)^(1/2)} / 2
n = 3 +or- 2(2)^(1/2)

set n equal to e^x
3 +or- 2(2)^(1/2) = e^x

e^x = 3 + 2(2)^(1/2) or e^x = 3 - 2(2)^(1/2)

log(e^x)=log[3 + 2(2)^(1/2)] or log(e^x)=log[3 - 2(2)^(1/2)]

Answer:
x = log [3 + 2(2)^(1/2)] or x = log [3 - 2(2)^(1/2)]

Answer 2

e^x + e^-x would not be e^0, that would be e^x * e^-x

e^-x = 1/e^x, so multiplying through by e^x on both sides would give you (e^x * e^x + e^x / e^x)/2 = 3e^x.

from here, e^x * e^x = e^2x, e^x / e^x = 1, and I'd multiply both sides by 2 to clear out the fraction, simplifying to e^2x + 1 = 6e^x

This equation is now in the form of au^2 + bu + c = 0 if you subtract 6e^x from both sides. Say u = e^x. Then, u^2 - 6u +1 = 0. This can be solved for u using the quadratic formula: (-b +/- sqrt(b^2 - 4ac)) / 2a = u. (+/- is plus or minus)

Using the quadratic formula, you get u = 3 +/- sqrt(8) = 3 +/- 2sqrt(2). Because we said u = e^x, we can substitute in and get e^x = 3 +/- 2sqrt(2). To solve this for x, take the natural log (ln) of both sides. This gives ln(e^x) = ln(3 +/- 2sqrt(2)). Solving this for x gives you x = ln(3 +/- 2sqrt(2)).

I hope you have some way of finding the ln(3 +/- 2sqrt(2)), but that's the exact answer.

Answer 3

No, to MULTIPLY you add exponents. You are adding. It would be e^2x/e^x +1/e^x = 6, or e^2x+1 = 6e^x - you'd get to that also if you followed the hint in the book to multiply each side by e^x.

Answer 4

14

Answer 5

(e^x + e^-x)/2 = 3

(e^x + e^-x) = 2(3) =6
multiply each term by e^x to get
e^(2x) +1 = 6e^x
e^(2x) -6e^x +1 =0

let t = e^x
t^2 -6t +1 =0
solve this

Discriminant = 6^2 -4 = 32
t1 = (6 - √32)/2 = 3 - 2√2
t2 = 3 + 2√2

now solve for x
t =e^x ----> x = ln(t)
x1 = ln(0.1715) = -1.763
x2 = ln(5.828) = 1.7627

Extra:
(e^x + e^-x)/2 = cosh(x) =3
x = cosh^(-1) of 3

Answer 6

i think 14 but im not sure cause i know u wouldnt listen to an 8th grader even if they were in adv. placement math =(

Answer 7

14 im a math major

Answer 8

17?

Answer 9

i fell bad for u :S

Answer 10

huh?