ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: tolga on December 30, 2009, 05:50:45 pm

Title: exponential equations
Post by: tolga on December 30, 2009, 05:50:45 pm: sove for x
C:\Users\TEMP\Pictures\Untitled.jpg
Title: Re: exponential equations
Post by: stonecold on December 30, 2009, 05:53:48 pm: Can you write it normally, or copy the pic from the text?
Title: Re: exponential equations
Post by: kriptik on December 30, 2009, 06:11:53 pm: wheres x in the picture?
Title: Re: exponential equations
Post by: brightsky on December 30, 2009, 06:13:45 pm: Out of curiosity, how do you solve this equation (assuming that he meant this):

$1.386 = x log_e2$
Title: Re: exponential equations
Post by: brightsky on December 30, 2009, 06:15:33 pm: Also, how do you solve log squared?

Say:

$log^2_2 x = 5$
Title: Re: exponential equations
Post by: Damo17 on December 30, 2009, 06:16:51 pm: Quote from: brightsky on December 30, 2009, 06:13:45 pm
Out of curiosity, how do you solve this equation (assuming that he meant this):

$1.386 = x log_e2$

$1.386=x \ log_e 2$

$\frac{1.386}{x}=log_e 2$

$\therefore$ $x=\frac{1.386}{log_e 2}$
Title: Re: exponential equations
Post by: brightsky on December 30, 2009, 06:18:11 pm: Ah I see. Cheers. ;)
Title: Re: exponential equations
Post by: Stroodle on December 30, 2009, 06:28:03 pm: Quote from: brightsky on December 30, 2009, 06:15:33 pm
Also, how do you solve log squared?

Say:

$log^2_2 x = 5$

$(log_2x)^2=5$

$log_2x=\pm\sqrt{5}$

$x=2^{\sqrt{5}}$ or $x=2^{-\sqrt{5}}$
Title: Re: exponential equations
Post by: brightsky on December 30, 2009, 06:30:51 pm: Ah thanks. But I don't think that would work with the equation on tolga's other post..

$log_2 (x^4) = log^2_2 x$

I've only done one step:

$4 log_2 x = log^2 _ 2 x$
Title: Re: exponential equations
Post by: Stroodle on December 30, 2009, 06:38:17 pm: I've never actually solved an equation with a squared log, but I would say:

$log_2x^4=(log_2x)^2$

$4=\frac{(log_2x)(log_2x)}{log_2x}$

$4=log_2x$

$\therefore x=16$

But this must be missing a solution, and I'm too tired to think right now...
Title: Re: exponential equations
Post by: brightsky on December 30, 2009, 06:44:52 pm: and x = 1 I think.

Thanks!! Was getting all tangled up with that one. :)
Title: Re: exponential equations
Post by: tolga on December 30, 2009, 07:27:17 pm: how did you get x=1
Title: Re: exponential equations
Post by: Stroodle on December 30, 2009, 07:48:58 pm: $log_2x^4=(log_2x)^2$

$(log_2x)^2-4log_2x=0$

$log_2x(log_2x-4)=0$

$log_2x=0\ \ or\ \ log_2x=4$

$\therefore x=1\ \ or\ \ x=16$