Exponential equation?!!?!? Help please!?

Question

Okay, here we go! I have NO IDEA how to solve this! Everything that I try yields the wrong answer!

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

Please show me step-by-step how you're supposed to solve that! I'll pick a best answer TODAY! Thank you so much!

Answer 1

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

(e^x + e^-x) = 3 (e^x - e^-x) -->

e^x + e^-x = 3e^x - 3e^-x -->

4e^-x = 2e^x -->

2e^-x = e^x -->

ln(2e^-x) = ln(e^x)

-------------------------------
Remember:

ln(a * b) = ln(a) + ln(b)
ln(e^x) = x * ln(e); ln(e^-x) = -x*ln(e)
ln(e) = 1

Apply to the formula above and you get:
---------------------------------

ln(2e^-x) = ln(e^x) -->

ln(2) - x*ln(e) = x*ln(e) -->

ln(2) - x = x -->

2x = ln(2) -->

x = ln(2)/2


Hope this helps :-)

Answer 2

Hmmm.... you arent going to like it....

The entire left side of the equation is equivalent to:
the hyperbolic cotangent (coth, or hypcot) of x

coth x = 3

x = arccoth 3
x ≈ 0.3465735902799...


Lets take an alternate approach:

(e^x + e^-x)/(e^x - e^-x) = 3
e^x + e^-x = 3•(e^x - e^-x)
e^x + e^-x = 3•e^x - 3•e^-x
4•e^-x = 2•e^x
2•e^-x = e^x
2 = (e^x)²
2 = e^(2x)
ln 2 = 2x
½•ln 2 = x

x = ½ • ln 2
x ≈ 0.3465735902799...

In case youre wondering, both ways come to the same conclusion.
½ • ln 2 = arccoth 3

Answer 3

Multiply both sides by e^x - e^(-x), and we have

e^x + e^(-x) = 3 e^x - 3 e^(-x).

Now add - e^x - e^(-x) to both sides, giving us

0 = 2 e^x - 4 e^(-x).

Next, add 4 e^(-x) to both sides.

4 e^(-x) = 2 e ^x.

Now, multiply both sides by 1/2 e^(x), resulting in

2 = e^(2x).

Taking the natural log of both sides, we have

ln(2) = 2x.

Diving both sides by two, and flipping the equation gives us

x = ln(2)/2.

There may or not be other solutions, I'm not sure, but I believe from the form of the equation that there should not be.

Answer 4

(e^x + e^-x)/(e^x - e^-x) = 3
e^x + e^-x = 3 * (e^x - e^-x) ; multiply by (e^x - e^-x)
e^(2x) + 1 = 3 * [e^(2x) - 1)] ; multiply by e^x
e^(2x) + 1 = 3e^(2x) - 3 ; Simplify right side
4 = 2e^(2x) ; add -e^(2x) + 3
2 = e^(2x) ; divide by 2
ln 2 = 2x ; take logarithm
1/2 * ln 2 = x

Answer 5

tanh^-1(1/3) = 0.3465736

Answer 6

cosh(x) = (e^x + e^-x) / 2
sinh(x) = (e^x - e^-x) / 2

coth = cosh(x) / sinh(x) = (e^x + e^-x)/(e^x - e^-x)
in your case
you would have

coth(x) = 3
and i believe everyone else has the same numerical answer so just use one of them

Answer 7

you can not solve this analyticly, you have to be satisfied with an approximation, for instance with newton's method.

Answer 8

do you have a scientific calculator? you know they are built in to your pc, but it looks like x=3