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General Discussion => Other Discussions => Math and Science => Topic started by: aeTIos on November 21, 2012, 05:53:23 am

Title: Solving this
Post by: aeTIos on November 21, 2012, 05:53:23 am

So yeah, I had a math exam today and the last question was this: f(x) = x*2^(2x+3), there is a horizontal line y=q that intersects the y axis in A and f(x) in B and C with AB:BC = 1:2. Calculate q in 2 decimals.
graph:
(http://img.removedfromgame.com/imgs/fx=x2^2x+3.bmp)
I didn't finish it during the exam (too bad) but this is what I came at afterwards:
f(-x) = f(-3x)  ; final answer is in -x, for why take a look at the graph
-x*2^(-2x+3)=-3x*2^(-6x+3)
-x*2^(-2x)*8=-3x*2^((-2x)^3)*8
-x*2^(-2x)=-3x*2^((-2x)^3)
;;substitute p for 2^(-2x)
-xp=-3xp^3
3p^2=1
p^2=1/3
p=sqrt(1/3)
2^(-2x) = sqrt(1/3)
log2(sqrt(1/3))=-2x
-x = log2(sqrt(sqrt(1/3)))
now input log2(sqrt(sqrt(1/3))) in f and you get y=~-1.83 so q = -1.83
Did I solve this right? Are there faster or better ways to do it?

Title: Re: Solving this
Post by: jacobly on November 21, 2012, 06:21:37 am

-x 2^-2x+3=-3x 2^-6x+3
2^-2x+3/2^-6x+3=-3x/-x
2^{(-2x+3)-(-6x+3)}=3 ; x!=0
log₂(3)=(-2x+3)-(-6x+3)
log₂(3)=4x
x=log₂(3)/4

Title: Re: Solving this
Post by: aeTIos on November 21, 2012, 06:24:57 am

Ah yeah I see, thanks :D
Actually what I did was in fact the same but with more steps, especially substitution. (which I use a lot since I can't see 2x^x as one number D:)

Title: Re: Solving this
Post by: ElementCoder on December 20, 2012, 04:06:56 am

I don't even understand anymore why to solve f(-x)=f(3x) O_O
[edit] sorry for the kind-of necropost :P