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Year 13 Maths Help Thread

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Original post by B_9710
I think you should probably form a quadratic equation in e^x and go from there.
Doing it this way you may struggle to solve e2x−ex−1=0 e^{2x}-e^x-1=0 .


But there is no 1 in the original equation?
Original post by DylanJ42
looks fine;

now just find the value of x for which x=2x \displaystyle x = 2x



So zero? :dontknow:
(edited 7 years ago)
Original post by kiiten
So zero? :dontknow:


perfect :biggrin:

as a matter of fact, going on what B_9710 said, how would you solve e2x−ex−1=0[br] \displaystyle e^{2x}-e^x-1=0[br] (just to make sure you know the method as you could be tested on it)
(edited 7 years ago)
Original post by kiiten
So zero? :dontknow:




:dots::dontknow::dontknow::dontknow::dontknow::dontknow::dontknow:


Yes, it is 0

What B_9710 was trying to say was you'd struggle to solve a similar equation involving a number, such as the one in his example, with the method you used. So if you knew how to solve it as a quadratic equation in ex it would be to your benefit.
Original post by DylanJ42
perfect :biggrin:


Thanks :smile: - i got confused because i did

x=2x
0 = 2x /x
0 = 2 ????

Is that wrong??

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Original post by kiiten
Thanks :smile: - i got confused because i did

x=2x
0 = 2x /x
0 = 2 ????

Is that wrong??

Posted from TSR Mobile


yea because x = 0 and so dividing by x creates a whole world of problems

from x = 2x just bring everything to the right of the equals sign(or the left, it doesnt matter) and get 0 = x
Original post by DylanJ42
yea because x = 0 and so dividing by x creates a whole world of problems

from x = 2x just bring everything to the right of the equals sign(or the left, it doesnt matter) and get 0 = x


2x / x isnt x though? (Bringing everything to the right of the equation)

Posted from TSR Mobile
Different ques

???? I got y = 2 / ln

Where did i go wrong?

Posted from TSR Mobile
1475414727151.jpg34.8KB
1475414769926.jpg28.7KB
Original post by kiiten
2x / x isnt x though? (Bringing everything to the right of the equation)

Posted from TSR Mobile


sorry I worded that awfully, i mean "-x" from both sides of the equation to get 0 = x
Original post by kiiten
Different ques

???? I got y = 2 / ln

Where did i go wrong?

Posted from TSR Mobile


log(a+b+c)≠loga+logb+logclog(a+b+c) \neq loga + logb + logc

For your other question, your equation is right - just solve it. If you had gone on to solve it and then put it back into the original question you would have been right - and you could have validated that for yourself rather than asking. Do not be afraid to try :h:
(edited 7 years ago)
Original post by kiiten
Different ques

???? I got y = 2 / ln

Where did i go wrong?

Posted from TSR Mobile


first off, ln⁡(a+b+c)≠ln⁡(a)+ln⁡(b)+ln⁡(c) \displaystyle \ln(a + b + c) \neq \ln(a) + \ln(b) + \ln(c)

also you cant say ln(y)=2  ⟹  y=2ln \displaystyle ln(y) = 2 \implies y = \frac{2}{ln}

that is similar to saying f(x)=4  ⟹  x=4f \displaystyle f(x) = 4 \implies x = \frac{4}{f} , it just doesnt make any sense.

I think the easiest way to do this question is to realise x and y must take the form en \displaystyle e^n (where n is just a number). So write x=ea \displaystyle x = e^a , y=eb \displaystyle y = e^b and go from there.

If not take equation 2 and "e" both sides (ive no idea what this step is actually called) to get e(Left hand side)=e(Right hand side) \displaystyle e^{(\text{Left hand side})} = e^{(\text{Right hand side})} then get in terms in x and sub into equation 1, then work out the quadratic which will follow

Edit: whoops you had the answer already, i didnt even notice :getmecoat:
(edited 7 years ago)
Original post by SeanFM
log(a+b+c)≠loga+logb+logclog(a+b+c) \neq loga + logb + logc

For your other question, your equation is right - just solve it. If you had gone on to solve it and then put it back into the original question you would have been right - and you could have validated that for yourself rather than asking. Do not be afraid to try :h:


Original post by DylanJ42
first off, ln⁡(a+b+c)≠ln⁡(a)+ln⁡(b)+ln⁡(c) \displaystyle \ln(a + b + c) \neq \ln(a) + \ln(b) + \ln(c)

also you cant say ln(y)=2  ⟹  y=2ln \displaystyle ln(y) = 2 \implies y = \frac{2}{ln}

that is similar to saying f(x)=4  ⟹  x=4f \displaystyle f(x) = 4 \implies x = \frac{4}{f} , it just doesnt make any sense.

I think the easiest way to do this question is to realise x and y must take the form en \displaystyle e^n (where n is just a number). So write x=ea \displaystyle x = e^a , y=eb \displaystyle y = e^b and go from there.

If not take equation 2 and "e" both sides (ive no idea what this step is actually called) to get e(Left hand side)=e(Right hand side) \displaystyle e^{(\text{Left hand side})} = e^{(\text{Right hand side})} then get in terms in x and sub into equation 1, then work out the quadratic which will follow

Edit: whoops you had the answer already, i didnt even notice :getmecoat:


I dont already have the answer? unless you mean some of my working is correct.

Im confused - i rearranged equation 1 to get x. Then i subbed that into the 2nd equation to get

ln (e^5 - e^2 + y) + lny = 7
Now ? - do you bring lny to the other side then divide by ln?
Original post by kiiten
I dont already have the answer? unless you mean some of my working is correct.

Im confused - i rearranged equation 1 to get x. Then i subbed that into the 2nd equation to get

ln (e^5 - e^2 + y) + lny = 7
Now ? - do you bring lny to the other side then divide by ln?


You can't can't can't divide by ln on its own - as Dylan has already said in his post. lne^3 = 3 does not imply that e^3 = 3 / ln(....) because ln(...) without an argument is meaningless. Just like how you wouldn't divide sinx + cosx by sin(nothing) - it would be meaningless.

As Dylan has taken time to give you hints I will not proceed any further as it may be confusing to have input from multiple people. Good luck!
Original post by SeanFM

As Dylan has taken time to give you hints I will not proceed any further as it may be confusing to have input from multiple people. Good luck!


I was on the reply screen as you replied but you were first, it was my bad. :sorry:
Original post by kiiten
I dont already have the answer? unless you mean some of my working is correct.

Im confused - i rearranged equation 1 to get x. Then i subbed that into the 2nd equation to get

ln (e^5 - e^2 + y) + lny = 7
Now ? - do you bring lny to the other side then divide by ln?


look on the right hand side of your page, you have ln⁡y=2 \displaystyle \ln y = 2 . How do you get y on its own here?

Edit: wait i think you fluked getting this right :laugh:
(edited 7 years ago)
Original post by DylanJ42
look on the right hand side of your page, you have ln⁡y=2 \displaystyle \ln y = 2 . How do you get y on its own here?


Ohhh so ....
e^lny = e^2
y = e^2
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 ln⁡(a+b+c)≠ln⁡(a)+ln⁡(b)+ln⁡(c) \displaystyle \displaystyle \ln(a + b + c) \neq \ln(a) + \ln(b) + \ln(c) but yet youve used it in your working, and it only didnt matter because y=e2 \displaystyle y = e^2

Think of it like having to find x when x=22 \displaystyle x = 2^2 , if you do x=2×2 \displaystyle x = 2 \times 2 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/showpost.php?p=67855350&postcount=875.

Another way is to get to ln⁡(e5−e2+y)+ln⁡y=7 \displaystyle \ln(e^5 - e^2 + y) + \ln y = 7 then "e" both sides to get;

eln⁡(e5−e2+y)+ln⁡y=e7 \displaystyle e^{\ln(e^5 - e^2 + y) + \ln y} = e^7

you'll then get a quadratic to solve
Original post by DylanJ42
yea thats the answer but im 99% you would lose marks because ln⁡(a+b+c)≠ln⁡(a)+ln⁡(b)+ln⁡(c) \displaystyle \displaystyle \ln(a + b + c) \neq \ln(a) + \ln(b) + \ln(c) but yet youve used it in your working, and it only didnt matter because y=e2 \displaystyle y = e^2

Think of it like having to find x when x=22 \displaystyle x = 2^2 , if you do x=2×2 \displaystyle x = 2 \times 2 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/showpost.php?p=67855350&postcount=875.

Another way is to get to ln⁡(e5−e2+y)+ln⁡y=7 \displaystyle \ln(e^5 - e^2 + y) + \ln y = 7 then "e" both sides to get;

eln⁡(e5−e2+y)+ln⁡y=e7 \displaystyle e^{\ln(e^5 - e^2 + y) + \ln y} = e^7

you'll then get a quadratic to solve


Thanks :smile:

Like this?

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

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1475439652995.jpg38.3KB
Original post by kiiten
Thanks :smile:

Like this?

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

Posted from TSR Mobile


Anyone? :s-smilie:

Posted from TSR Mobile
Original post by kiiten


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

Remember that xa+b=xa×xbx^{a + b} = x^a \times x^b .

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