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FP3 Super Quick Log Question!

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I get that the rationalised the denominator but how does ln(1- (1-x etc etc = -ln(1+root1 etc etc

basically I don't understand the final line
Reply 1
Original post by creativebuzz
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I get that the rationalised the denominator but how does ln(1- (1-x etc etc = -ln(1+root1 etc etc

basically I don't understand the final line


1(1x2)=11+x2=x21-(1-x^2) = 1-1 + x^2 = x^2

So 1(1x2)x(1+1x2)=x2x(1+1x2)=x1+1x2\displaystyle \frac{1-(1-x^2)}{x(1+ \sqrt{1-x^2})} = \frac{x^2}{x(1 + \sqrt{1-x^2})} = \frac{x}{1 + \sqrt{1-x^2}}

Now you have ln(x1+1x2)\displaystyle \ln \left( \frac{x}{1 + \sqrt{1-x^2}}\right)

And you know that lna=ln1a\ln a = -\ln \frac{1}{a}

So ln1+1x2x=lnx1+1x2\displaystyle -\ln \frac{1 + \sqrt{1-x^2}}{x} = \ln \frac{x}{1 + \sqrt{1-x^2}}
Original post by Zacken
1(1x2)=11+x2=x21-(1-x^2) = 1-1 + x^2 = x^2

So 1(1x2)x(1+1x2)=x2x(1+1x2)=x1+1x2\displaystyle \frac{1-(1-x^2)}{x(1+ \sqrt{1-x^2})} = \frac{x^2}{x(1 + \sqrt{1-x^2})} = \frac{x}{1 + \sqrt{1-x^2}}

Now you have ln(x1+1x2)\displaystyle \ln \left( \frac{x}{1 + \sqrt{1-x^2}}\right)

And you know that lna=ln1a\ln a = -\ln \frac{1}{a}

So ln1+1x2x=lnx1+1x2\displaystyle -\ln \frac{1 + \sqrt{1-x^2}}{x} = \ln \frac{x}{1 + \sqrt{1-x^2}}


Ah I see! Last question, would you mind spotting where I went wrong my working out

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Reply 3
Original post by creativebuzz
Ah I see! Last question, would you mind spotting where I went wrong my working out


Have you posted the right picture? I genuinely have no clue what's going on.

You want to solve the system;

x+2y+z=2x+2y+z = 2 and 6x+y4z=166x + y - 4z = 16, you decided to let y=λy = \lambda. This gives you two equations in two variables:

x+z=22λx +z = 2 - 2\lambda and 6z4z=16λ6z - 4z = 16 - \lambda, you can now solve for xx and zz in terms of λ\lambda.
Original post by Zacken
Have you posted the right picture? I genuinely have no clue what's going on.

You want to solve the system;

x+2y+z=2x+2y+z = 2 and 6x+y4z=166x + y - 4z = 16, you decided to let y=λy = \lambda. This gives you two equations in two variables:

x+z=22λx +z = 2 - 2\lambda and 6z4z=16λ6z - 4z = 16 - \lambda, you can now solve for xx and zz in terms of λ\lambda.


Oops sorry ignore the first picture, the last two are the relevant ones!

(I was meant to send

x + 2y + z = 2

6x + y - 4x =16

6x + 12y + 6z = 12
6x + y - 4z = 16

11y + 10z = -4

and then the rest of my working out is in the last picture I sent you)
Reply 5
Original post by creativebuzz
Oops sorry ignore the first picture, the last two are the relevant ones!

(I was meant to send

x + 2y + z = 2

6x + y - 4x =16

6x + 12y + 6z = 12
6x + y - 4z = 16

11y + 10z = -4

and then the rest of my working out is in the last picture I sent you)


Why are you solving for lambda? Solve for x and z in terms of lambda.

I can't give a more detailed reply because I'm not home, sorry!
Original post by Zacken
Why are you solving for lambda? Solve for x and z in terms of lambda.

I can't give a more detailed reply because I'm not home, sorry!


Didn't I do that in the rest of my working out?

Also, quick question, how do you do find the mod arg form of modz = root3

Untitled (2).png

I know that the mod is root3 but I can't seem to get the argument
Reply 7
Original post by creativebuzz
Didn't I do that in the rest of my working out?

Also, quick question, how do you do find the mod arg form of modz = root3

Untitled (2).png

I know that the mod is root3 but I can't seem to get the argument


It says argw=7π12\arg w = \frac{7\pi}{12} right on top. arg\arg means argument. So the argument of ww is 7π12\frac{7\pi}{12}.
Original post by Zacken
It says argw=7π12\arg w = \frac{7\pi}{12} right on top. arg\arg means argument. So the argument of ww is 7π12\frac{7\pi}{12}.


o
m
g:facepalm:
Reply 9
Original post by creativebuzz
o
m
g:facepalm:


Perhaps taking a break and relaxing for a while is in order? :tongue:
Original post by Zacken
Perhaps taking a break and relaxing for a while is in order? :tongue:

Lol yup!

Do you know how I can draw the graph V = 0.5tsin8t?
Reply 11
Original post by creativebuzz
Lol yup!

Do you know how I can draw the graph V = 0.5tsin8t?


Well, starts at (0,0) - has roots at 8t=π,2π,3π,t=π8,2π8,3π8,8t = \pi, 2\pi, 3\pi, \cdots \Rightarrow t = \frac{\pi}{8}, \frac{2\pi}{8}, \frac{3\pi}{8}, \cdots bascally every integer multiple of π8\frac{\pi}{8}. The amplitude gets bigger and bigger but it always bounded between the lines y=±0.5ty = \pm 0.5t, maxima and minima exactly where you would expect them. That should be enough to draw the graph.

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