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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: ahmed on March 17, 2009, 10:37:27 pm

Title: Spec. problems-need help asap
Post by: ahmed on March 17, 2009, 10:37:27 pm: Need help with the following questions:(@=theta)
1. Convert the following to the standard form $Rcos(@-B)$
a) 7cos @ + 7root3sin @ b) 4cos @ - 4root3sin @
2.Simplify
a) 4 ln(2x^2y) - 7ln(xy) + 4ln(y)
3.Solve for x:
a) y = 3 - 2ln((x-5)/2)
b) 8y + 3= 2e^(1 + 2x)
4.Find the real numbers (x) and (y) if (x - iy)(1+5i) - 3i(x + iy) = (2 - i)^2 - 7i

Thankz for the help in advance.
Title: Re: Spec. problems-need help asap
Post by: Damo17 on March 17, 2009, 10:59:22 pm: 1a
$7 cos \theta +7\sqrt{3} sin \theta$

Find R which is $\sqrt{a^2+b^2}$ where $a=7$ and $b=7\sqrt{3}$
$R=\sqrt{7^2+(7\sqrt{3})^2}$
$= 14<br />$
$tan \theta=\frac{b}{a}$
$tan \theta= \frac{7\sqrt{3}}{7}$
$\therefore \theta= tan^{-1}(\sqrt{3})$
$\theta= 1.047$

$\therefore$ $7 cos \theta +7\sqrt{3} sin \theta= 14cos(x-1.047)$
Title: Re: Spec. problems-need help asap
Post by: dcc on March 18, 2009, 10:26:57 pm: Quote from: ahmed on March 17, 2009, 10:37:27 pm
2. Simplify
$4 ln(2x^{2}y) - 7ln(xy) + 4ln(y)$

$A = 4 \ln(2x^2) + 4\ln(y) - 7\ln(x) - 7\ln(y) + 4\ln(y)$
$A = 4 \ln(2) + \ln(x) + \ln(y)$
Title: Re: Spec. problems-need help asap
Post by: ahmed on March 18, 2009, 11:18:13 pm: Quote from: dcc on March 18, 2009, 10:26:57 pm
Quote from: ahmed on March 17, 2009, 10:37:27 pm
2. Simplify
$4 ln(2x^{2}y) - 7ln(xy) + 4ln(y)$

$A = 4 \ln(2x^2) + 4\ln(y) - 7\ln(x) - 7\ln(y) + 4\ln(y)$
$A = 4 \ln(2) + \ln(x) + \ln(y)$

how did u get this answer, $A = 4 \ln(2) + \ln(x) + \ln(y)$ could you explain it.
Thankz for all ur help.
Title: Re: Spec. problems-need help asap
Post by: TonyHem on March 18, 2009, 11:35:34 pm: 3
a) $y = 3 - 2ln((x-5)/2)$
$0 = 3-2ln((x-5)/2)$
$3/2 = ln((x-5)/2)$
$(x-5)/2 = E^{3/2}$
$x = 2E^{3/2}+5$
$x = 13.96$

b) $8y+3 = 2e^(1+2x)$
$y=0, 3/2 = e^{1+2x}$
$1+2x = loge 3/2$
$x = (loge 3/2)-1/2$ <-- not sure how to put this in latex. Loge 3/2 -1 all over 2.
$x= -0.297$

edit: For that log question above, he just broke it up into parts.
The $4Ln(2x^2y)$ broken into $4ln(2) + 4ln(x^2) + 4ln(y)$
$-7n(xy) into -7ln(x)-7ln(y)$
$4ln(y) -> ln(y^4$ ) ltr

$4ln(2x^2y)-7ln(xy)+4ln(y)$
$4ln(2)+4ln(x^2)+4ln(y)-(7ln(x)+7ln(y))+4ln(y)$
$4ln(2)+ln(x^8)-ln(x^7)+ln(y^4)+ln(y^4)-ln(y^7)$ <-----alogex = logex^a and logex+logex = loge(x^2)
$4ln(2)+ln(x)+ln(y)$

$4.Find the real numbers (x) and (y) if (x - iy)(1+5i) - 3i(x + iy) = (2 - i)^2 - 7i$

Expand the brackets, simplify it and then group the real and imaginary parts together.
Then let the real part which should be $x+8y = 3$ ( on the left side) and let the imaginary part $(2x-y) = -11$
So simultaneous equations
I got $y = 1 x = -5$

$x+5xi-yi-5yi^2 -3xi-3yi^2 = 4-4i+i^2-7i$
$x+8y+2xi-yi = 3-11i$
$x+8y = 3$ [times by 2] --[1]
$2x-y = -11 [2]$
$[1]$ x 2 $= 2x+16y = 6 -- [3]$
$[3]-[2]$
$17y = 17$
$y = 1$
$sub y=1 into 2, 2x-1 = -11$
$x = -5$

ps: half of those q's r MM ones