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Reply 40
Original post by King-Panther

How would I differentiate 1/x^2


Differentiate x2 \displaystyle x^{-2}
Reply 41
Original post by King-Panther
So why would I plot logy against log x, I don't understand?

Yeah, I did and I got x^2 + 2x + 1/x^2

How would I differentiate 1/x^2


Because logy is dependant on log x so in other words you plot logy against logx. What this ends up meaning is, your log y values are on the vertical co-ordinate axis, whilst your logx values are on the horizontal co-ordinate axis in the cartesian plane.

1/x^2 is the same as x^-2.
Original post by f1mad
Because logy is dependant on log x so in other words you plot logy against logx. What this ends up meaning is, your log y values are on the vertical co-ordinate axis, whilst your logx values are on the horizontal co-ordinate axis in the cartesian plane.

1/x^2 is the same as x^-2.


Original post by raheem94
Differentiate x2 \displaystyle x^{-2}



Thank you both!!!!
Original post by f1mad
Because logy is dependant on log x so in other words you plot logy against logx. What this ends up meaning is, your log y values are on the vertical co-ordinate axis, whilst your logx values are on the horizontal co-ordinate axis in the cartesian plane.

1/x^2 is the same as x^-2.


Im stuck on 20, I've used sn=1/2n(a+L), but im unsure what to do

DW, I got it, thanks anyway!
(edited 11 years ago)
Original post by King-Panther


The original equation is (x/1)-(1/x1/2)dx



I don't understand WHY you have written (x/1) surely you should have written x
Reply 45
Original post by King-Panther
Im stuck on 20, I've used sn=1/2n(a+L), but im unsure what to do



Technically, there are infinite number of possibilities. Although if you let a= 2, 3, 4 in turn and let the common difference be d.

You can use s100 = 1000 to find d, and hence allowing you to find an nth term.

Oops, I see you got it :tongue:.
Original post by f1mad
Technically, there are infinite number of possibilities. Although if you let a= 2, 3, 4 in turn and let the common difference be d.

You can use s100 = 1000 to find d, and hence allowing you to find an nth term.

Oops, I see you got it :tongue:.


Thats fine, I've got a mac, Q23) b looks like -f(x), is that right?

For a, I got points (x+3)^2
Reply 47
Original post by King-Panther
Thats fine, I've got a mac, Q23) b looks like -f(x), is that right?

For a, I got points (x+3)^2


It is -f(x), so what does this end up giving?

The curve is (x+3)^2 yes, now you need to draw it.
Original post by steve2005
I don't understand WHY you have written (x/1) surely you should have written x

I was trying to make it simpler. Can you give me a hand on 24 & 25 please?
Original post by f1mad
It is -f(x), so what does this end up giving?

The curve is (x+3)^2 yes, now you need to draw it.


For a, its a curve crossing the y at 9 and touching x at -3.... For b, its touching (0,0) but the over way round...

For 26, do I use logs

i've got up to log3x+1=2
(edited 11 years ago)
Original post by King-Panther
I was trying to make it simpler. Can you give me a hand on 24 & 25 please?





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Original post by steve2005



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So yes for 25, still unsure about 24.... Thanks
Original post by King-Panther
So yes for 25, still unsure about 24.... Thanks


24. yes it is true, the first wiki image steve2005 put up states that if f(x)<0 f''(x)<0 then f has a local maximum at x.
i.e. if the second derivative of f(x) is less than 0 (negative), there is a local maximum.
So clearly all local maximum points will produce a negative second derivative :smile:
Original post by just george
24. yes it is true, the first wiki image steve2005 put up states that if f(x)<0 f''(x)<0 then f has a local maximum at x.
i.e. if the second derivative of f(x) is less than 0 (negative), there is a local maximum.
So clearly all local maximum points will produce a negative second derivative :smile:


thanks....... Any advice for 26 & 27
(edited 11 years ago)
Original post by King-Panther
thanks....... Any advice for 26 & 27



Well for 26. try writing 4 as (2×2) (2\times2) so you have (2×2)x(2\times2)^x
how can you then write that as 2 to the power of something?
And then from there, can you solve it using logs (base 2)?
Reply 55
Original post by King-Panther
thanks....... Any advice for 26 & 27


Question 27 is, log(x)+log(2x)log(x3)=log0.5 \displaystyle log(x) + log(2x) -log(x^3)=log0.5

Remember, loga+logblogc=logablogc=log(abc) \displaystyle loga + logb - logc = logab - logc = log\left(\frac{ab}{c}\right)

So log(abc)=log0.5    abc=0.5 \displaystyle log\left(\frac{ab}{c}\right) = log0.5 \implies \frac{ab}{c}=0.5
Original post by just george
Well for 26. try writing 4 as (2×2) (2\times2) so you have (2×2)x(2\times2)^x
how can you then write that as 2 to the power of something?
And then from there, can you solve it using logs (base 2)?


So I'll have 3x+1log2=2xlog2
Reply 57
Original post by King-Panther
thanks....... Any advice for 26 & 27


For question 26,
23x+1=4x \displaystyle 2^{3x+1}=4^{x}
Take log of both sides,
log(23x+1)=log(4x)    (3x+1)log2=xlog4 \displaystyle log\left(2^{3x+1}\right)=log \left(4^{x}\right) \implies (3x+1)log2=xlog4
Original post by raheem94
Question 27 is, log(x)+log(2x)log(x3)=log0.5 \displaystyle log(x) + log(2x) -log(x^3)=log0.5

Remember, loga+logblogc=logablogc=log(abc) \displaystyle loga + logb - logc = logab - logc = log\left(\frac{ab}{c}\right)

So log(abc)=log0.5    abc=0.5 \displaystyle log\left(\frac{ab}{c}\right) = log0.5 \implies \frac{ab}{c}=0.5


so,log(abc)=log(x(2x)x3)=log(2x2x3) so, \displaystyle log\left(\frac{ab}{c}\right) = \displaystyle log\left(\frac{x(2x)}{x^3}\right) = \displaystyle log\left(\frac{2x^2}{x^3}\right)

???
Reply 59
Original post by King-Panther
log(x(2x)x3)=log(2x2)x3) \displaystyle log\left(\frac{x(2x)}{x^3}\right) = \displaystyle log\left(\frac{2x^2)}{x^3}\right)

???


log(2x2x3)    log(2x) \displaystyle log\left(\frac{2x^2}{x^3}\right) \implies log\left(\frac2{x}\right)

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