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# Year 13 Maths Help Thread

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1. (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
Remember that is asking "what numbers can I square to get 4?" And the answer is obviously , but don't forget that also works.

In general, if you have then your solutions are , so can you apply this here?
2. (Original post by RDKGames)
That is worrying if you're confused on that. Remember that , 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 cannot have x as negatives. When it comes to however, then you CAN have negative x values because negatives/positives squared just become positive.

When you go from to 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).
3. Fs.
It's just x^2=|x|^2

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4. (Original post by kiiten)
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).
When you have then can't be negative, since you can't have a negative number inside the logarithm.

When you have then can be negative, since will still be positive anyway and you can have that inside the logarithm.
5. (Original post by kiiten)
Kinda. Please could you remind me of that core 2 example you were talking about (ive forgotten most of it :3 )
Here's an example from Core 2. Solve:

This of course reads as "what is the number I get, x, when I raise the base, 2, to the third power?"

From here you would fully go with cancelling the logarithm by making both sides an index of a certain number, that number being the base of the logarithm:

and you should know that therefore the equation becomes:

Now you can apply the same idea to your equation without moving the exponent of x down.

(Original post by kiiten)
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).
No, there ARE negative 'x' values but you are getting rid off them with the way you are working through your answer by moving the exponent of x down as then you have and x cannot be negative for this expression, while x can be negative in as negative numbers squared become positive, so you'd be taking the log of a positive number which is valid hence negative x values are valid.
6. How can I start to find in terms of ?

That is, just drop me a hint and not the full solution.
7. (Original post by Palette)
How can I start to find in terms of ?

That is, just drop me a hint and not the full solution.
Differentiate the geometric series.
8. (Original post by Zacken)
Differentiate the geometric series.
The derivative of gives but I don't know how it will help.

Would this differentiation approach work for any where g(r) is a geometric series?
9. (Original post by Palette)
The derivative of gives but I don't know how it will help.

Would this differentiation approach work for any where g(r) is a geometric series?
Not what I meant.

or along these lines w/e

and no to your last question
10. (Original post by Zacken)
Not what I meant.

or along these lines w/e

and no to your last question
I don't want to sound stupid, but why can we differentiate when can only take positive integer values between [0 and ?
11. (Original post by Palette)
I don't want to sound stupid, but why can we differentiate when can only take positive integer values between [0 and ?
Simplified, your question is isomorphic to: "Why can we differentiate when is a positive integer?"

I explicitly pointed out that we're differentiating with respect to (a variable that can vary over the reals) by writing and not . We're not differentiating with respect to , it's just a constant.
12. (Original post by Zacken)
Simplified, your question is isomorphic to: "Why can we differentiate when is a positive integer?"

I explicitly pointed out that we're differentiating with respect to (a variable that can vary over the reals) by writing and not . We're not differentiating with respect to , it's just a constant.
I get it now- thanks for your help!
13. (Original post by Zacken)
Simplified, your question is isomorphic to: "Why can we differentiate when is a positive integer?"

I explicitly pointed out that we're differentiating with respect to (a variable that can vary over the reals) by writing and not . We're not differentiating with respect to , it's just a constant.
Having looked at my question again, I realized how silly it was.
14. (Original post by Palette)
Having looked at my question again, I realized how silly it was.
The original one or the "r constant" one?

By the way, even if could take on positive integer values, you could still take on as a real variable, differentiate with respect to it and then send it back to the integer space in nice enough conditions and/or sufficient handwaving.
15. (Original post by Palette)
Having looked at my question again, I realized how silly it was.
Same - i only realise when its pointed out to me. Im just glad im anonymous on this site aha

- thanks to the people who helped me on that ln question. I think i understand now
16. (Original post by k.russell)
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)
Yeah that makes sense thanks a lot, forgot to include a=0 when it's at constant velocity ie terminal velocity
17. (Original post by metrize)
Yeah that makes sense thanks a lot, forgot to include a=0 when it's at constant velocity ie terminal velocity

lol np, I am glad (and surprised) I could help out I am guessing you are studying some M units, how many have you done so far?
18. Sorry to intrude, but just out of interest, I'd like to ask those who do both physics and further maths which is the more difficult? (I do further maths, not physics)

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19. (Original post by RedSquirrels)
Sorry to intrude, but just out of interest, I'd like to ask those who do both physics and further maths which is the more difficult? (I do further maths, not physics)

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Further maths can be very easy if you're interested in maths.
20. (Original post by RedSquirrels)
Sorry to intrude, but just out of interest, I'd like to ask those who do both physics and further maths which is the more difficult? (I do further maths, not physics)

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I think Physics. You can choose fairly 'easy' (as in work hard, practice a lot, gain understanding and ace the exam) modules and it can be almost... straightforward.

Physics doesn't test your mathematical abilities as much but there's a lot of theory, used to have oracticals as well which was a nightnare, and like a lot of science subjects the exams/markschemes are quite annoying about things.

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