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# Year 13 Maths Help Thread

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1. (Original post by kiiten)
Ohhh so ....
e^lny = e^2
y = e^2
yea thats the answer but im 99% you would lose marks because but yet youve used it in your working, and it only didnt matter because

Think of it like having to find x when , if you do you still get the right answer but you're working it wrong and you've just fluked it, that's kind of similar to whats happened here.

My previous post includes 2 ways of doing it, http://www.thestudentroom.co.uk/show...&postcount=875.

Another way is to get to then "e" both sides to get;

you'll then get a quadratic to solve
2. (Original post by DylanJ42)
yea thats the answer but im 99% you would lose marks because but yet youve used it in your working, and it only didnt matter because

Think of it like having to find x when , if you do you still get the right answer but you're working it wrong and you've just fluked it, that's kind of similar to whats happened here.

My previous post includes 2 ways of doing it, http://www.thestudentroom.co.uk/show...&postcount=875.

Another way is to get to then "e" both sides to get;

you'll then get a quadratic to solve
Thanks

Like this?

If this is right then im not sure how you would factorise after this.

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Attached Images

3. (Original post by kiiten)
Thanks

Like this?

If this is right then im not sure how you would factorise after this.

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Anyone?

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4. Your first half of the LHS is right but you have brought e^...+ lny down to give lny, which isn't quite right

Remember that .
5. (Original post by SeanFM)
Your first half of the LHS is right but you have brought e^...+ lny down to give lny, which isn't quite right

Remember that .
What do you mean i havent brought down lny??
6. (Original post by kiiten)
What do you mean i havent brought down lny??
Okay, how did you get from the line with e^ln(...) to the line below? how did you get the +lny in the equation below it?
7. (Original post by SeanFM)
Okay, how did you get from the line with e^ln(...) to the line below? how did you get the +lny in the equation below it?
I moved everything after e^ln in front of e^ln (e^5 - e^2 etc.). Then e^ln cancels out which leaves e^5 - e^2 etc.
lny is in the equation because that was also part of the index notation?
8. (Original post by kiiten)
Different ques

???? I got y = 2 / ln

Where did i go wrong?

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I'm not quite sure what you are trying to do but there is a whole lot simpler way to solve it if you get rid off the natural logarithm.

Just use the fact that

So

Then rearrange for, say, and we get

Subbing it into the first equation we get:

Which leads to and just solve for and then .
9. (Original post by kiiten)
I moved everything after e^ln in front of e^ln (e^5 - e^2 etc.). Then e^ln cancels out which leaves e^5 - e^2 etc.
lny is in the equation because that was also part of the index notation?
the e^5 - e^2 - y can be taken down but notice that it adds lny...

What you've done isn't completely correct is what I'm trying to say.

where a and b are whatever you want.
10. (Original post by RDKGames)
I'm not quite sure what you are trying to do but there is a whole lot simpler way to solve it if you get rid off the natural logarithm.

Just use the fact that

So

Then rearrange for, say, and we get

Subbing it into the first equation we get:

Which leads to and just solve for and then .
Thanks - i understand what youre saying here but im not too sure how you would solve to find y.
EDIT: sorry its in pencil :3
11. (Original post by SeanFM)
the e^5 - e^2 - y can be taken down but notice that it adds lny...

What you've done isn't completely correct is what I'm trying to say.

where a and b are whatever you want.
So what happens to the lny? Do you mean it doesnt work because lny isnt inside the bracket?
12. (Original post by kiiten)
So what happens to the lny? Do you mean it doesnt work because lny isnt inside the bracket?
That is what I am trying to show you (what happens to the lny)

Like I said from the previous post,

In this case a is ln(e^5 - e^2 + y) and b is ln(y).

So e^a+b is e^(ln(e^5 - e^2 + y) * e^(lny) = ....

Now you should be able to see the difference between what you did in post 882 and what has just been done.
13. (Original post by SeanFM)
That is what I am trying to show you (what happens to the lny)

Like I said from the previous post,

In this case a is ln(e^5 - e^2 + y) and b is ln(y).

So e^a+b is e^(ln(e^5 - e^2 + y) * e^(lny) = ....

Now you should be able to see the difference between what you did in post 882 and what has just been done.
Ahh i see. Thank youu - sorry i just wasnt getting it at all but i understand now. Just to check does this lead to:

y(e^5 - e^2 + y) = e^7
y = e^7 / (e^5 - e^2 + y)
y = e^2 - e^5 + y/(e^5 - e^2 + y)

??
14. (Original post by kiiten)
Ahh i see. Thank youu - sorry i just wasnt getting it at all but i understand now. Just to check does this lead to:

y(e^5 - e^2 + y) = e^7
y = e^7 / (e^5 - e^2 + y)
y = e^2 - e^5 + y/(e^5 - e^2 + y)

??
After your first line of the quoted post it is easier to turn it into a quadratic, as you can see you've reached a dead end.. I think.
15. (Original post by SeanFM)
After your first line of the quoted post it is easier to turn it into a quadratic, as you can see you've reached a dead end.. I think.
Yep . Ohh So:

ye^5 - ye^2 + y^2 = e^7
y^2 - ye^2 + ye^5 - e^7 = 0
y(y - e^2) + e^5 (y - e^2) = 0
(y+ e^5)(y - e^2) = 0
y = e^2

Finally, yay - is this right? - question only asks for +ve solutions so the other bracket is invalid.
16. (Original post by kiiten)
Yep . Ohh So:

ye^5 - ye^2 + y^2 = e^7
y^2 - ye^2 + ye^5 - e^7 = 0
y(y - e^2) + e^5 (y - e^2) = 0
(y+ e^5)(y - e^2) = 0
y = e^2

Finally, yay - is this right? - question only asks for +ve solutions so the other bracket is invalid.
You can always check the solutions of an equation by putting them back into the equation
17. (Original post by SeanFM)
You can always check the solutions of an equation by putting them back into the equation
Yea I think i figured it out at last. The solutions work. Thanks so much for your help

One more question - dy/dx of (x+1)^3(x+2)^6 do you use the chain rule?
I ended up with 3(x+1)^2 6(x+2)^5 but the question wants it in the form (x+1)^2 (x+2)^5 (ax+b) ??
18. (Original post by kiiten)
Yea I think i figured it out at last. The solutions work. Thanks so much for your help

One more question - dy/dx of (x+1)^3(x+2)^6 do you use the chain rule?
I ended up with 3(x+1)^2 6(x+2)^5 but the question wants it in the form (x+1)^2 (x+2)^5 (ax+b) ??
Think about the product rule too
19. (Original post by SeanFM)
Think about the product rule too
Ah but im not sure about using the product rule with brackets. I tried it with (x+1)^3 and got:

(x+1)3(x+1)^2 + (x+1)^3 ?
20. (Original post by kiiten)
Ah but im not sure about using the product rule with brackets. I tried it with (x+1)^3 and got:

(x+1)3(x+1)^2 + (x+1)^3 ?
The product rule is when you have two functions multiplying eachother - you're mixing it up with the chain rule a bit, though you do have to (spoiler alert) use the chain rule here.

If you say f(x) = (x+1)^3 and g(x) = (x+2)^6.. you do the rest

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