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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: wolfmanalpha on June 16, 2013, 08:52:14 pm

Title: how do i solve this by hand ?
Post by: wolfmanalpha on June 16, 2013, 08:52:14 pm: how do i solve for x by hand:

1/2(e^x-e^-x) = -3/4

answer is -log(2)

thanks for any help
Title: Re: how do i solve this by hand ?
Post by: Daenerys Targaryen on June 16, 2013, 09:06:20 pm: $\frac{1}{2}(e^x-e^{-x})=\frac{-3}{4}$

$(e^x-e^{-x})=\frac{-3}{2}$

Let $e^x=a$

$a-a^{-1}=\frac{-3}{2}$

$a^2-1=\frac{-3}{2} a$

$2a^2+3a-2=0$

$(2a-1)(a+2)=0$

$\implies e^x=\frac{1}{2}$ or $\implies e^x=-2$ But not possible

$\therefore log_e(\frac{1}{2})=x$

$log_e(1)-log_e(2)$

$-log_e(2)$
Title: Re: how do i solve this by hand ?
Post by: Alwin on June 16, 2013, 09:06:42 pm: Quote from: wolfmanalpha on June 16, 2013, 08:52:14 pm
how do i solve for x by hand:

1/2(e^x-e^-x) = -3/4

answer is -log(2)

thanks for any help

$\frac{1}{2}(e^x-e^{-x}) = -\frac{3}{4}$

let a dummy variable be $u=e^x$ Hence:

$\frac{1}{2}(u-\frac{1}{u}) = -\frac{3}{4}$
$u-\frac{1}{u}= -\frac{3}{2}$

Times both sides by $u$ ,
$u^2-1=-\frac{3}{2}u$
$2u^2+3u-2=0$
$(2u-1)(u+2)=0$

From the null factor law, this gives
$u=\frac{1}{2}, \text{ or } u=-2$
$\therefore e^{x}=\frac{1}{2} e^{x}=-2$ since we let u=e^x at the start
$\therefore e^{x}=\frac{1}{2}$ since e^x > 0, we reject the second solution

Hence:
$x=\ln{\frac{1}{2}}$
$x=-\ln{2}$ as required

EDIT: beaten
Title: Re: how do i solve this by hand ?
Post by: wolfmanalpha on June 16, 2013, 09:07:29 pm: thanks for the help guys
Title: Re: how do i solve this by hand ?
Post by: BubbleWrapMan on June 16, 2013, 09:32:17 pm: There's this if you're interested. Particularly the hyperbolic sine and its inverse, which was essentially used here. Of course, the methods used in this thread are far more intuitive, but nonetheless:

https://en.wikipedia.org/wiki/Hyperbolic_function