The Student Room Group

Quick Log Questions

Question 2logx=log(x-1)+log3

So using the laws of log, I rearranged to:

logx^2=log((x-1)3)

therefore

x^2 = 3x-3
x^2 -3x+3=0

However there are no real answers? Is this correct?

Thank you.
(edited 11 years ago)

Scroll to see replies

Reply 1
what happened to the -2 going to 2 in the exponent?
Original post by danny111
what happened to the -2 going to 2 in the exponent?


Sorry, it was just meant to be 2, not -2. That was a typo.
Original post by danny111
does b24acb^2 - 4ac mean anything to you?


Yes, I realise there would be two imaginary roots however I find that unlikely as it's a different maths module. Is all of my workings above correct? If they are then I can confidently answer with the imaginary roots.

Thanks a lot for your help btw :smile:
Reply 4
Original post by Jerking My Gerking
Yes, I realise there would be two imaginary roots however I find that unlikely as it's a different maths module. Is all of my workings above correct? If they are then I can confidently answer with the imaginary roots.

Thanks a lot for your help btw :smile:


Your working is correct, I don't understand why just because complex numbers are a different unit they shouldn't come up in another?
Do you have any idea with these other 3 questions I'm stuck with:

Solve:

1. 6cosx=4sinx

2. 4(sin)^2 x+8cosx=7

3. 3sin^3 x-5sin^2 x-6sinx+8
for this question I've already done part be, where I expressed 3s^3-5s^2-6s+8 as (s-1)(s-2)(3s+4) and solved to s=1, s=2 and s=-4/3.
However, I don't understand how this will help me solve this part.

Any help with any of the above is greatly appreciated.
Reply 6
When I was in sixth form I had a page with all the trig identities on them from the standard ones to more complicated ones. If I were you I would make one too and then these questions should be straightforward.
Reply 7
Original post by Jerking My Gerking


3. 3sin^3 x-5sin^2 x-6sinx+8
for this question I've already done part be, where I expressed 3s^3-5s^2-6s+8 as (s-1)(s-2)(3s+4) and solved to s=1, s=2 and s=-4/3.
However, I don't understand how this will help me solve this part.

Any help with any of the above is greatly appreciated.


Try writing s = sin x and see if you recognize anything!
Reply 8
Original post by danny111
Your working is correct, I don't understand why just because complex numbers are a different unit they shouldn't come up in another?


Because most students completing C3 will not be doing FP1
Reply 9
Original post by danny111
When I was in sixth form I had a page with all the trig identities on them from the standard ones to more complicated ones. If I were you I would make one too and then these questions should be straightforward.


Or save a lot of hassle and remember them perhaps?
Reply 10
Original post by TenOfThem
Because most students completing C3 will not be doing FP1


You can do A level maths without ever being introduced to complex numbers? That's just lame.
Reply 11
Original post by Madalaine M
Or save a lot of hassle and remember them perhaps?


Really :s-smilie: never thought of that...
Original post by danny111
You can do A level maths without ever being introduced to complex numbers? That's just lame.


ok
Reply 13
Original post by TenOfThem
ok


You don't think students embarking on maths at sixth form should be made aware of the cool thing that imaginary numbers are?
Original post by danny111
You don't think students embarking on maths at sixth form should be made aware of the cool thing that imaginary numbers are?


Sure

But I spent the first dozen years of my teaching careers re-writing the scheme of work for A Level, year on year removing more and more content

It is not about what I think is ok

It is about the reality
Reply 15
Original post by TenOfThem
Sure

But I spent the first dozen years of my teaching careers re-writing the scheme of work for A Level, year on year removing more and more content

It is not about what I think is ok

It is about the reality


What does that mean exactly?

And you sound really, well bitter may be strong word, but disillusioned(?) by the system?
Original post by Jerking My Gerking
Question 2logx=log(x-1)+log3

So using the laws of log, I rearranged to:

logx^2=log((x-1)3)

therefore

x^2 = 3x-3
x^2 -3x+3=0

However there are no real answers? Is this correct?

Thank you.


I think you have probably made an error in typing out the question.
Original post by TenOfThem
Sure

But I spent the first dozen years of my teaching careers re-writing the scheme of work for A Level, year on year removing more and more content

It is not about what I think is ok

It is about the reality


Ah! Perhaps this brings back some good memories TenofThem...

Cambridge_GCE_Mathematics_1960.pdf

Some real beasts on there!
Original post by Madalaine M
Ah! Perhaps this brings back some good memories TenofThem...

Cambridge_GCE_Mathematics_1960.pdf

Some real beasts on there!


Cheeky

I was only 50 last week :biggrin:

1981 was my year
Original post by danny111
What does that mean exactly?

And you sound really, well bitter may be strong word, but disillusioned(?) by the system?


The system is what it is

If I were too disillusioned I would not have been a part of the system for the last 28 years

Quick Reply

Latest