Solve for x in 3 - e^x - e^-x = 0?

Question

Apparently you need to let e^x = u
Then solve it as a quadratic using the quadratic formula
But I can't see how!

Answer 1

Hey there!

If you let u=e^x, then e^-x=(e^x)^-1 or u^-1.

So you have 3-u-u^-1=0.

Here's the answer.

3-u-u^-1=0 --> Write the problem.
-u-u^-1=-3 --> Subtract 3 on both sides of the equation.
u+u^-1=3 --> Divide both sides of the equation by -1.
u^2+1=3u --> Multiply both sides of the equation by u.
u^2-3u+1=0 --> Subtract 3u on both sides of the equation.

Now apply the quadratic formula.

u=3±sqrt(9-4(1)(1))/(2)(1) -->
u=3±sqrt(9-4)/2 -->
u=3±sqrt(5)/2

Now substitute back into the problem.

u=3±sqrt(5)/2 --> Write the answer.
e^x=3±sqrt(5)/2 --> Substitute e^x for u.
x=ln((3±sqrt(5))/2) Take ln on both sides of the equation.

So the answer is x=ln((3±sqrt(5))/2). Since ln((3-sqrt(5))/2) gives us an undefined solution, it is eliminated from the solutions.

Correct answer is x=ln((3+sqrt(5))/2).

Hope it helps!

Answer 2

3 - e^x - e^-x = 0

Multiply by e^x:

3e^x - (e^x)^2 - 1 = 0

Now rearrange and multiply through by minus 1:

(e^x)^2 - 3e^x + 1 = 0

There's your quadratic. To make things easier to handle, you can let e^x = u as you said.

u^2 - 3u + 1 = 0

Using the standard method for solving a quadratic ( I presume you know it from algebra) you get a negative and a positive answer. The negative gives an irrational, so we'll drop that one.

The solution is:

u = e^x = 2.618

x = ln e^x = 0.9624

The answer is correct. I substituted into the original equation

Answer 3

Let u = e^x

3 -e^x -e^-x = 3-e^x - 1/e^x = 3 - u -1/u =0

Multiply by u:

3u - u^2 - 1 = 0 or

u^2 - 3u -1 = 0

u = 1/2*3 +/-1/2*sqrt(3^2 -4*(-1)*(1)) = 3/2+/-1/2*sqrt(9+4)

u = 3/2 +/- 1/2*sqrt(13)

There are two roots - one corresponding to the "+" sign and one corresponding to the "-" in the above equation. Solving;

u = 3.303, -0.3023

Since u = e^x then x = ln(u) (natural log)

so x = 1.195 corresponding to u = 3.303.

There is no value of x that makes e^x negative, so the second value of u has no corresponding value of x

Answer 4

First amplify this inequality to grant x^2 -x -3x +3 > 0 Then simplify to grant: x^2 -4x +3 >0 Then draw a graph. to help, use the unique equation to get the coordinates the place x=0: (0,3) and (0,a million) as quickly as the graph is drawn, you will discover that x is barely under (or equivalent to) 0 between the standards y=3 and y=a million.. hence written as an inequality: x>3 or x

Answer 5

3 - e^x - e^-x = 0
3 - u - 1/u = 0
3u - u^2 - 1 = 0
u^2 - 3u + 1 = 0
u = (3 +- sqrt(9 - 4))/2
u = (3 +- sqrt(5))/2
either u = 2.618 or u = 0.38

e^x = 2.618
x = ln(2.618) = 0.96

e^x = 0.38
x = ln(0.38) = -0.96

Answer 6

x=ln[(3+sqrt. of 5)/2]

Answer 7

e^-x = 1/e^x

3 - e^x - 1/e^x = 0

ok, so u = e^x

3 - u - 1/u = 0
3u - u^2 - 1 = 0

-1 (u^2 - 3u + 1) = 0

use quadratic formula:
-b +/- sqrt(b^2 - 4ac)
---------------------------
............. 2a

3 +/- sqrt(3^2 - 4(1)(1))
------------------------------
.............. 2(1)


3 +/- sqrt(5)
----------------
...... 2


e^x = [ 3 + qrt(5)]/2
x = ln [ 3 + qrt(5)]/2

e^x = [3 - sqrt(5)]/2
x = ln [3 - sqrt(5)]/2


so the answer is:
x =~ +/- 0.96242

Answer 8

Notice that
cosh(x)=(1/2)*(e^x + e^(-x))
This changes your equation to
3-2cosh(x)=0
x=arccosh(3/2)
x=0.9624...