# P3 Algebra Questions

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Hi, I'm stuck with several things.

1. v = ln (2x-3) dv/dx=2/2x-3

I thought ln x = 1/x so ln 2x-3 = 1/2x-3... Why?? Please somebody explain this.

If you could give me some clues on how to tackle 2 & 3, would help so much.

2. Given that lxl <1, expand [(1+x)/(1-x)]^(1/3) in ascending powers of x and up to and including the term in x^2.

3. The coefficients of x and x^2 in the expansion of (1+px+qx^2)^-2 in ascending pwoers of x are 4 and 14 respectively. Find the values of p and q.

Thank you!!!

1. v = ln (2x-3) dv/dx=2/2x-3

I thought ln x = 1/x so ln 2x-3 = 1/2x-3... Why?? Please somebody explain this.

If you could give me some clues on how to tackle 2 & 3, would help so much.

2. Given that lxl <1, expand [(1+x)/(1-x)]^(1/3) in ascending powers of x and up to and including the term in x^2.

3. The coefficients of x and x^2 in the expansion of (1+px+qx^2)^-2 in ascending pwoers of x are 4 and 14 respectively. Find the values of p and q.

Thank you!!!

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#2

(Original post by

Hi, I'm stuck with several things.

1. v = ln (2x-3) dv/dx=2/2x-3

I thought ln x = 1/x so ln 2x-3 = 1/2x-3... Why?? Please somebody explain this.

Thank you!!!

**TheQueen1986**)Hi, I'm stuck with several things.

1. v = ln (2x-3) dv/dx=2/2x-3

I thought ln x = 1/x so ln 2x-3 = 1/2x-3... Why?? Please somebody explain this.

Thank you!!!

I think..it's been awhile since i did maths

G

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#3

(Original post by

Hi, I'm stuck with several things.

1. v = ln (2x-3) dv/dx=2/2x-3

I thought ln x = 1/x so ln 2x-3 = 1/2x-3... Why?? Please somebody explain this.

If you could give me some clues on how to tackle 2 & 3, would help so much.

2. Given that lxl <1, expand [(1+x)/(1-x)]^(1/3) in ascending powers of x and up to and including the term in x^2.

3. The coefficients of x and x^2 in the expansion of (1+px+qx^2)^-2 in ascending pwoers of x are 4 and 14 respectively. Find the values of p and q.

Thank you!!!

**TheQueen1986**)Hi, I'm stuck with several things.

1. v = ln (2x-3) dv/dx=2/2x-3

I thought ln x = 1/x so ln 2x-3 = 1/2x-3... Why?? Please somebody explain this.

If you could give me some clues on how to tackle 2 & 3, would help so much.

2. Given that lxl <1, expand [(1+x)/(1-x)]^(1/3) in ascending powers of x and up to and including the term in x^2.

3. The coefficients of x and x^2 in the expansion of (1+px+qx^2)^-2 in ascending pwoers of x are 4 and 14 respectively. Find the values of p and q.

Thank you!!!

Nevertheless if y = ln x then dy/dx = 1/x, that may be where your confusion came from. So if y = ln (2x-3) then we apply the chain rule i.e. y = ln t, t=2x-3

then dy/dx = 1/t * dt/dx and dt/dx = 2 so dy/dx = 2/t = 2/(2x-3)

2. Clue: Taylor series, take the first few derivatives and evaluate at 0.

3. Again you need to use Taylor series, take a few derivatives treating p and q as constants and it should follow.

If you need more help just ask or pm me.

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#4

(Original post by

ln x is the natural logarithm of x. If you are having problem with a basic function like the logarithm function then you need to go and revise quite a bit.

Nevertheless if y = ln x then dy/dx = 1/x, that may be where your confusion came from. So if y = ln (2x-3) then we apply the chain rule i.e. y = ln t, t=2x-3

then dy/dx = 1/t * dt/dx and dt/dx = 2 so dy/dx = 2/t = 2/(2x-3)

2. Clue: Taylor series, take the first few derivatives and evaluate at 0.

3. Again you need to use Taylor series, take a few derivatives treating p and q as constants and it should follow.

If you need more help just ask or pm me.

**AntiMagicMan**)ln x is the natural logarithm of x. If you are having problem with a basic function like the logarithm function then you need to go and revise quite a bit.

Nevertheless if y = ln x then dy/dx = 1/x, that may be where your confusion came from. So if y = ln (2x-3) then we apply the chain rule i.e. y = ln t, t=2x-3

then dy/dx = 1/t * dt/dx and dt/dx = 2 so dy/dx = 2/t = 2/(2x-3)

2. Clue: Taylor series, take the first few derivatives and evaluate at 0.

3. Again you need to use Taylor series, take a few derivatives treating p and q as constants and it should follow.

If you need more help just ask or pm me.

taylor series is not taught in P3 only binomial expansion can be used for this

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OK thanks, I get about the logarithm stuff.

But for 2 and 3, I don't know the Taylor series, never been taught that. Does anyone know a binomial way of approaching?? Thanks.

But for 2 and 3, I don't know the Taylor series, never been taught that. Does anyone know a binomial way of approaching?? Thanks.

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#7

(Original post by

OK thanks, I get about the logarithm stuff.

But for 2 and 3, I don't know the Taylor series, never been taught that. Does anyone know a binomial way of approaching?? Thanks.

**TheQueen1986**)OK thanks, I get about the logarithm stuff.

But for 2 and 3, I don't know the Taylor series, never been taught that. Does anyone know a binomial way of approaching?? Thanks.

Then you can use binomial expansion to figure it out..i think

G

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#8

**TheQueen1986**)

Hi, I'm stuck with several things.

1. v = ln (2x-3) dv/dx=2/2x-3

I thought ln x = 1/x so ln 2x-3 = 1/2x-3... Why?? Please somebody explain this.

If you could give me some clues on how to tackle 2 & 3, would help so much.

2. Given that lxl <1, expand [(1+x)/(1-x)]^(1/3) in ascending powers of x and up to and including the term in x^2.

3. The coefficients of x and x^2 in the expansion of (1+px+qx^2)^-2 in ascending pwoers of x are 4 and 14 respectively. Find the values of p and q.

Thank you!!!

[(1+x)^(1/3)][(1-x)^(-1/3)]

Then expand the two separately and multiply them.

If you need help on the actual expansion then PM me.

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#9

Q3:

(1+ px+qx^2)^-2.

Put px + qx^2 = t. Using binomial expansion:

(1 + t)^-2 = 1 + (-2)t + (-2)(-3)t^2/2! + ...

= 1 + (-2)(px+qx^2) + 3(px +qx^2)^2 + ...

= 1 + (-2p)x + (-2q + 3p^2)x^2 +...

So you got the power of x and x^2. Replace then solve.

Try this.

(1+ px+qx^2)^-2.

Put px + qx^2 = t. Using binomial expansion:

(1 + t)^-2 = 1 + (-2)t + (-2)(-3)t^2/2! + ...

= 1 + (-2)(px+qx^2) + 3(px +qx^2)^2 + ...

= 1 + (-2p)x + (-2q + 3p^2)x^2 +...

So you got the power of x and x^2. Replace then solve.

Try this.

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Sorry, still have some probs.

Q2:

I did what you did, but I still have troubles.

Ok I expand the numerator and the denominator. But to what x^??

Do you mean when you multiply the two, that you multiply numerator by 1/denominator? So if the expanded denominator was something like (just making it up for now) 1+2x+3x^2 you make it 1+2x^-1+3x^-2.

Q3:

How did you get from:= 1 + (-2)(px+qx^2) + 3(px +qx^2)^2 + ...

----> 1 + (-2p)x + (-2q + 3p^2)x^2 +...

Q2:

I did what you did, but I still have troubles.

Ok I expand the numerator and the denominator. But to what x^??

Do you mean when you multiply the two, that you multiply numerator by 1/denominator? So if the expanded denominator was something like (just making it up for now) 1+2x+3x^2 you make it 1+2x^-1+3x^-2.

Q3:

How did you get from:= 1 + (-2)(px+qx^2) + 3(px +qx^2)^2 + ...

----> 1 + (-2p)x + (-2q + 3p^2)x^2 +...

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#11

(Original post by

2. Given that lxl <1, expand [(1+x)/(1-x)]^(1/3) in ascending powers of x and up to and including the term in x^2.

**TheQueen1986**)2. Given that lxl <1, expand [(1+x)/(1-x)]^(1/3) in ascending powers of x and up to and including the term in x^2.

[(1+x)/(1-x)]^(1/3) = [(1+x)^(1/3)][(1-x)^(-1/3)]

Now using binomial expansiona dn slittping the two expressions seperately:

(1+x)^(1/3) = 1 + x/3 + [(1/3)(-2/3)x^2]/2 + ...

(1-x)^(-1/3) = 1 + x/3 + [(-1/3)(-4/3)x^2]/2 + ...

Now multiplying the two expanded expressions and collecting the terms only upto x^2

[(1+x)/(1-x)]^(1/3) = [1 + x/3 + (x^2)/9 + ...][1 + x/3 + (2/9)(x^2) + ...]

= 1 +x/3 + (2/9)(x^2) + x/3 + (1/9)(x^2) + (1/9)(x^2)

= 1 + (2/3)x + (4/9)(x^2)

Therefore the answer is:

**[(1+x)/(1-x)]^(1/3) = 1 + (2/3)x + (4/9)(x^2) + ...**

================================ ==========================

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#13

(Original post by

Sorry, still have some probs.

Q2:

I did what you did, but I still have troubles.

Ok I expand the numerator and the denominator. But to what x^??

Do you mean when you multiply the two, that you multiply numerator by 1/denominator? So if the expanded denominator was something like (just making it up for now) 1+2x+3x^2 you make it 1+2x^-1+3x^-2.

Q3:

How did you get from:= 1 + (-2)(px+qx^2) + 3(px +qx^2)^2 + ...

----> 1 + (-2p)x + (-2q + 3p^2)x^2 +...

**TheQueen1986**)Sorry, still have some probs.

Q2:

I did what you did, but I still have troubles.

Ok I expand the numerator and the denominator. But to what x^??

Do you mean when you multiply the two, that you multiply numerator by 1/denominator? So if the expanded denominator was something like (just making it up for now) 1+2x+3x^2 you make it 1+2x^-1+3x^-2.

Q3:

How did you get from:= 1 + (-2)(px+qx^2) + 3(px +qx^2)^2 + ...

----> 1 + (-2p)x + (-2q + 3p^2)x^2 +...

|px + qx^2| < 1. But in ur question, i didnt find that, maybe there's another way for this (not Taylor series).

But if it uses binomial expansion, as u know

(1 + x)^n = 1 + nx + n(n-1)x^2/2! + n(n-1)(n-2)x^3/3! + .....

But i think it's right, cuz it's P2 question.

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#15

A fellow A level student trying to finish their A2 maths before the year starts? Heh.

Q1: Use the chain rule, don't make false assumptions:

V = ln(2x-3)

V = ln(t), t=2x-3

dv/dt = 1/t, dt/dx =2

dv/dx = 2/t,

dv/dx = 2/(2x-3)

Q1: Use the chain rule, don't make false assumptions:

V = ln(2x-3)

V = ln(t), t=2x-3

dv/dt = 1/t, dt/dx =2

dv/dx = 2/t,

dv/dx = 2/(2x-3)

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X

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