natural logarithm/solve for x?

Question

e^x - e^(-x) = 1

uh oh

thanks you guys

Answer 1

Your best bet is trial and error. This can be rewritten as exp(x) - 1/exp(x) = 1. Clearly, x will be somewhat greater than "1".

Answer 2

Don't use "trial an error" or any other kind of guessing. You can actually solve this analytically.

Multiplying both sides by e^x gives
e^(x)e^(x) - e^(-x)e^(x) = e^(x)
(e^x)^2 - 1 = e^(x)
(e^x)^2 - (e^x) - 1 = 0

Let y = e^x, and you have y^2 - y - 1 = 0.
This solves to give you y = (1±√5)/2. This is e^x which has to be positive, so get rid of the negative solution of y to leave you with
e^x = ((1+√5)/2), which means
x = ln ((1+√5)/2)

Answer 3

do no longer use "trial an errors" or the different style of guessing. you may actual remedy this analytically. Multiplying the two aspects by utilising e^x supplies e^(x)e^(x) - e^(-x)e^(x) = e^(x) (e^x)^2 - a million = e^(x) (e^x)^2 - (e^x) - a million = 0 permit y = e^x, and you have y^2 - y - a million = 0. This solves to furnish you y = (a million±?5)/2. it extremely is e^x which should be helpful, so do away with the unfavorable answer of y to pass away you with e^x = ((a million+?5)/2), meaning x = ln ((a million+?5)/2)

Answer 4

a) just write e^x - 1/e^x = 1 and multiply each term by e^x

b) e^(2x) - 1 = e^x ==> e^(2x) - e^x - 1 = 0

c) Now change the variable: e^x = y and will get a quadratic equation... note that y must be positive number.

d) y^2 - y - 1 = 0 and the positive root is y= (1+sqrt(5))/2

e) Now take e^x = y and put the y obtained

e^x = (1+ sqrt(5))/2 and finally x = ln (1 + sqrt(5))/2)

Answer 5

I think the taylor series expansion e^x is
sum of (x^n/factorial(n)) from 0 to infinity and valid for all x so
when n is zero or even the terms in e^x and e^(-x) cancel each other and when n is odd they are the same so the difference expansion should be sum(2*x^n/factorial(n)) for n positive and odd and this is supposed to = 1.

Is this the same as saying sinh(x)=0.5?

Answer 6

let y=e^x
then y+1/y=1 or y^2-y-1=0
you will get one meaningful answer
and that is y = (1+√5)/2
and finally x=ln((1±√5)/2)=0.48
there's another way though
e^x-e^(-x)=1 then [e^x-e^(-x)]/2=1/2
sinh(x)=1/2 => x=arcsinh(1/2)=0.48
the same answer.

Answer 7

it is not too bad

You can say that e^x - e^(-x) = 2 sinh(x)

So:
sinh(x) = 1/2
arcsinh(1/2) = x
x = 0.48121

Also if you know some more about math you will know that

arcsinh(x) = ln(x + sqrt(x^2 + 1))
arcsinh(1/2) = ln(1/2 + sqrt(1/4 + 1)
arcsinh(1/2) = ln((1+sqrt(5))/2)

Answer 8

what kinds of questions do you ask? gigle will you save me.