ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Thymaster on March 16, 2013, 03:24:15 pm

Title: exponential query
Post by: Thymaster on March 16, 2013, 03:24:15 pm: 8e^-x -e^x=2 solve for x
Title: Re: exponential query
Post by: Daenerys Targaryen on March 16, 2013, 03:28:37 pm: $\frac{8}{e^x}-e^x=2$

Let $e^x=a$

$\frac{8}{a}-a=2$

$8-a^2=2a$

$a^2+2a-8=0$

$(a+4)(a-2)=0$

$a=-4$ or $a=2$

But $a=e^x$

$e^x=-4$ or $e^x=2$

$e^x=-4$ - not possibe

$x=log_{e}(2)$
Title: Re: exponential query
Post by: Thymaster on March 16, 2013, 03:46:43 pm: but why is it a^2
Title: Re: exponential query
Post by: Homer on March 16, 2013, 03:47:16 pm: Quote from: HatersGonnaHate on March 16, 2013, 03:28:37 pm

$8-a=2a$

should be $8-a^2=2a$

Quote from: Thymaster on March 16, 2013, 03:46:43 pm
but why is it a^2

$e^{2x}=(e^{x})^{2}$

hence $a^2$
Title: Re: exponential query
Post by: Thymaster on March 16, 2013, 03:50:53 pm: i mean isnt it meant to be a initially instead of a^2 since there in no 2 infornt of e^x
Title: Re: exponential query
Post by: Conic on March 16, 2013, 03:53:07 pm: Quote from: Thymaster on March 16, 2013, 03:46:43 pm
but why is it a^2
Multiplied $\frac{8}{a}-a=2$ by a which is $8-a^2=2a$ .
Title: Re: exponential query
Post by: Daenerys Targaryen on March 16, 2013, 03:53:22 pm: Woops typos ill edit it