Year 13 Maths Help Thread

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Why is the answer +- for this C3 question

ln x^2 = 8
2lnx = 8
lnx = 4
x= e^4

But it should be x= +- e^4
Why is it +- (im not taking the sq rt of 8?)

Reply 641

B_9710

Original post by kiiten

Why is the answer +- for this C3 question

ln x^2 = 8
2lnx = 8
lnx = 4
x= e^4

But it should be x= +- e^4
Why is it +- (im not taking the sq rt of 8?)

If you exponentiate both sides to begin with you get

x^2=e^8

Reply 642

RDKGames

Study Forum Helper

Original post by kiiten

Why is the answer +- for this C3 question

ln x^2 = 8
2lnx = 8
lnx = 4
x= e^4

But it should be x= +- e^4
Why is it +- (im not taking the sq rt of 8?)

You're not getting a negative because of your second step where ln(x) cannot take upon negative x values whereas in the first one the x in ln(x²) can indeed take upon both negative and positive ones. So technically speaking you can go ahead with your method as long as you remember to include the opposite sign of your answer.

(edited 7 years ago)

Reply 643

username1162972

Original post by kiiten

Why is the answer +- for this C3 question

ln x^2 = 8
2lnx = 8
lnx = 4
x= e^4

But it should be x= +- e^4
Why is it +- (im not taking the sq rt of 8?)

ln|x|=4 so |x|=e^4

Reply 644

youreanutter

Original post by SeanFM

Handy tip, at the end of a pdf link if you type '#page6' for example, it will take the reader to page 6 immediately.

OK do u know explanation for that answer

Reply 645

Kevin De Bruyne

Original post by youreanutter

OK do u know explanation for that answer

Please link me to the correct page (that is the reason I told you how to do it), I do not want to filter through a whole PDF to look for a particular reference.

Thank you :h:

Reply 646

asinghj

So i have this differentiation question,
Show that

\frac{d}{dx} ((4x-1)\sqrt{4x-1}) = 6\sqrt{4x-1}

I tried to use product rule, but the u and v are the same and i dont know how to go about it,
I tried the chain rule, but still have no clue, as there are x terms on both of them.
So basically i dont know where to get started

Reply 647

ValerieKR

Original post by asinghj

So i have this differentiation question,
Show that

\frac{d}{dx} ((4x-1)\sqrt{4x-1}) = 6\sqrt{4x-1}

product rule
u=sqrt(4x-1)
v=4x-1
derivative of uv = u*v' + u'*v

alternatively you're finding the derivative of (4x-1)^3/2, chain rule with 4x-1=u

Reply 648

asinghj

Original post by ValerieKR

product rule
u=sqrt(4x-1)
v=4x-1
derivative of uv = u*v' + u'*v

alternatively you're finding the derivative of (4x-1)^3/2, chain rule with 4x-1=u

OMG, thank you. I spent the whole day trying to figure this out and was overcomplicating so much, when the answer was so simple😂😂

Spoiler

Reply 649

metrize

can someone explain how to do 1d?
http://www.undergraduate.study.cam.ac.uk/files/publications/engaa_s2_specimen_question_paper.pdf

For part C i resolved the forces to make acceleration the subject and integrated it to get velocity, and I got D as the answer for that, not sure how to work out part D, do I have to resolve again and include kv as drag?

or do I do kv=ma, use the acceleration from previous section?

(edited 7 years ago)

Reply 650

kiiten

Original post by B_9710

If you exponentiate both sides to begin with you get

x^2=e^8

the question is ln x^2 = 8
im confused how you got from that to x^2 = e^8

Original post by RDKGames

Im confused :s-smilie:

Reply 651

k.russell

Hi gonna jump on this thread even though it's a bit late :smile:

I would like to be a helper and a learner lol as I have done C3, C4 and M2 and am now on a gap yahh doing FP1, FP2, S1 and S2. Did pretty good in both C3 and C4 so should be able to be quite helpful there, though maybe not as helpful as people who are really good at maths (i.e. those doing it at university)

Reply 652

k.russell

Original post by kiiten

the question is ln x^2 = 8
im confused how you got from that to x^2 = e^8

Im confused :s-smilie:

ln is log in base e, so you can do e to the power of both sides to eliminate the ln, understand?

Reply 653

k.russell

Original post by metrize

now I am a future biologist, so maybe not a leading authority on engineering - but I do think I've worked it out.
It makes sense the block will reach terminal velocity, right, when the forces acting forward = those acting backward.
So you set up the diagram, mg sin(a) down the slope and uN + kV up the slope, as we are adding in the air resistance term (kV). N = mg cos (a) , so you end up with this;
mg sin(a) = u mg cos(a) + kV
and you rearrange it get V out, because that V is gonna be the terminal velocity. At lesser values of V, the forces down slope would be greater and the object would accelerate. Do you agree with that?? Hope you can understand it lol (I had to use a for alpha and u for mu)

(edited 7 years ago)

Reply 654

RDKGames

Study Forum Helper

Original post by kiiten

the question is ln x^2 = 8
im confused how you got from that to x^2 = e^8

That is worrying if you're confused on that. Remember that

\ln(x^2)=\log_e(x^2)

, and you SHOULD be able to know what happens here after Core 2.

Original post by kiiten

Im confused :s-smilie:

Not sure how else to explain it to you. You should know that you cannot have a logarithm of a negative number, therefore

\ln(x)

cannot have x as negatives. When it comes to

\ln(x^2)

however, then you CAN have negative x values because negatives/positives squared just become positive.

When you go from

\ln(x^2)

2\ln(x)

you are essentially getting rid off all the negative x-values hence getting rid off one of the solutions. Get it?

Reply 655

kiiten

Original post by k.russell

ln is log in base e, so you can do e to the power of both sides to eliminate the ln, understand?

Ohhhh so you do e^(ln x^2) = e^8
X^2 = e^8
X = sqrt e^8

?? Wait ive gone wrong, sorry my minds just gone blank :s-smilie:

Reply 656

Zacken

Original post by kiiten

Ohhhh so you do e^(ln x^2) = e^8
X^2 = e^8
X = sqrt e^8

?? Wait ive gone wrong, sorry my minds just gone blank :s-smilie:

Remember that

x^2 = 4

is asking "what numbers can I square to get 4?" And the answer is obviously

x=2

, but don't forget that

x = -2 \Rightarrow x^2 = (-2)^2 = -2 \times -2 = 4

also works.

In general, if you have

x^2 = k

then your solutions are

x = \pm \sqrt{k}

, so can you apply this here?

Reply 657

kiiten

Original post by RDKGames

That is worrying if you're confused on that. Remember that

\ln(x^2)=\log_e(x^2)

, and you SHOULD be able to know what happens here after Core 2.

Not sure how else to explain it to you. You should know that you cannot have a logarithm of a negative number, therefore

\ln(x)

cannot have x as negatives. When it comes to

\ln(x^2)

however, then you CAN have negative x values because negatives/positives squared just become positive.

When you go from

\ln(x^2)

2\ln(x)

you are essentially getting rid off all the negative x-values hence getting rid off one of the solutions. Get it?

Kinda. Please could you remind me of that core 2 example you were talking about (ive forgotten most of it :3 )

Ok so if there are no -ve values why is the answet + or - ?? When it cant be a -ve? (I realised im over complicating it - sorry).