It's not taking too long, I just want to see if there is a faster way lol.(Original post by SeanFM)
Right.. and how have you solved them / why do you think its taking so much time? It could be that you're just getting used to it at the start.
I'm quite fast at calculating, so it's not too long tbh.
EDIT: Latex is being a pain in the ass, look at (g) I've used the same thing for (h) I'll put up in around 3 mins.
C2  Logarithms
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 23062016 16:48

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 23062016 16:54
(Original post by SeanFM)
Right.. and how have you solved them / why do you think its taking so much time? It could be that you're just getting used to it at the start. 
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 23062016 16:55
(Original post by Chittesh14)
Put up both methods. 
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 23062016 16:57
(Original post by Chittesh14)
Questions: Solve, giving your answer to 3 s.f.
(g) 3^{2x+1}  26(3^{x})  9 = 0
(h) 4(3^{2x+1}) + 17(3^{x})  7 = 0
How I solved them:
(g)
log_{3}9 = x
x = 2
When y = 1/3, there are no solutions.
(h)
log_{3}1/3 = x
x = 1
When y = 7/4, there are no solutions.
When you bring the 3 down in 3^(2x+1), leave it in front, so you get 3*(3^2x) and then substitute as normal. Otherwise, nothing else you can do to speed it up. 
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 23062016 17:00
(Original post by SeanFM)
I see why you feel like it's longer than it should be  you've divided by 3 and then multiplied by 3 again.
When you bring the 3 down in 3^(2x+1), leave it in front, so you get 3*(3^2x) and then substitute as normal. Otherwise, nothing else you can do to speed it up. 
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 23062016 17:06
(g)
Treat this as a quadratic expression, you can let u=3^{x }if you want wish to in order for it to appear simpler. Then factorise it.
Therefore or ; disregard as for all
> therefore from inspection (I guess you can use logs just to show it's true)
(h)
Treat it as a quadratic again and factorise it.
Therefore or ; disregard negative
> therefore from inspection
The only key thing here is to realise andLast edited by RDKGames; 23062016 at 17:26. Reason: Making use of latex 
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 24062016 15:49
(Original post by RDKGames)
(g)
Treat this as a quadratic expression, you can let u=3^{x }if you want wish to in order for it to appear simpler. Then factorise it.
Therefore or ; disregard as for all
> therefore from inspection (I guess you can use logs just to show it's true)
(h)
Treat it as a quadratic again and factorise it.
Therefore or ; disregard negative
> therefore from inspection
The only key thing here is to realise and
Thank you so much ♥, I'm going to do logs again later on today after I stop playing games!! 
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 25062016 18:49
(Original post by RDKGames)
x(Original post by SeanFM)
x
I know that you probably won't get what I'm saying lool, but I hope you do. 
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 25062016 18:52
(Original post by Chittesh14)
Does this statement: "You can divide a polynomial by " mean that you can divide a polynomial by (x (a positive or negative number) so e.g. (x2) or (x+2) right? I feel I misinterpreted it when I thought it meant you can divide any polynomial by p as a negative or positive number so again (x2) and (x+2).
I know that you probably won't get what I'm saying lool, but I hope you do.
If you fix a p value, whether it's positive or negative, the + /  will cover both ends.
But you're getting far too bogged down in the specification and such statements what's important is that you can divide by (x+a) or (xa) where a is a positive constant. 
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 25062016 19:07
(Original post by SeanFM)
I'm not entirely sure what the difference is between what you've said and what they've said.
If you fix a p value, whether it's positive or negative, the + /  will cover both ends.
But you're getting far too bogged down in the specification and such statements what's important is that you can divide by (x+a) or (xa) where a is a positive constant.
If a polynomial is let's say... x^3  10x^2 + 19x + 30, since (x+1) is a factor, I was thinking that if p = 1, (x1) is also a factor. But, that is incorrect because if you factorise it, you get: (x+1)(x5)(x6) lol.
I hope you understand what I am saying now . 
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 25062016 19:09
(Original post by Chittesh14)
Lol, no what I was thinking was:
If a polynomial is let's say... x^3  10x^2 + 19x + 30, since (x+1) is a factor, I was thinking that if p = 1, (x1) is also a factor. But, that is incorrect because if you factorise it, you get: (x+1)(x5)(x6) lol.
I hope you understand what I am saying now . 
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 25062016 19:33
(Original post by Chittesh14)
Does this statement: "You can divide a polynomial by " mean that you can divide a polynomial by (x (a positive or negative number) so e.g. (x2) or (x+2) right? I feel I misinterpreted it when I thought it meant you can divide any polynomial by p as a negative or positive number so again (x2) and (x+2).I know that you probably won't get what I'm saying lool, but I hope you do.(Original post by Chittesh14)
Lol, no what I was thinking was:If a polynomial is let's say... x^3  10x^2 + 19x + 30, since (x+1) is a factor, I was thinking that if p = 1, (x1) is also a factor. But, that is incorrect because if you factorise it, you get: (x+1)(x5)(x6) lol.I hope you understand what I am saying now .
The book is basically saying you CAN DIVIDE the polynomial by any linear expression such as however this does NOT mean and are BOTH roots of , therefore it does NOT mean and are both factors of . ONLY if gives a remainder of 0 will be a solution to it equaling zero. (or if divided through by )
Just because you can divide a polynomial by a linear expression, it does not strictly mean that linear expression has a root for the polynomial.
Practice long division with polynomial expressions to see what I mean.Last edited by RDKGames; 25062016 at 20:31. 
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 25062016 20:05
Just realised p(n) should be written as p(x) instead. Wouldn't make sense for a function of n to be expressed in terms of x, sorry. xD
But this is what I mean, therefore x=2 a real root if p(x)=0. If you plug in x=2, the first division will not be determined because you would be dividing by 0 on the denominator, thus x=2 cannot be a root.
Posted from TSR MobileLast edited by RDKGames; 25062016 at 20:30. 
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 25062016 20:27
In all honesty, I knew where I was wrong and what it meant. I just wanted to see you how both would react to it and explain it to me !
(Original post by SeanFM)
Ohh, I see what you mean. but yeah  just go with what I said about 'a' in my previous post.
(Original post by RDKGames)
I think I know what you mean and I see the flaw in your thought. Let's say you have some polynomial .
The book is basically saying you CAN DIVIDE the polynomial by any linear expression such as however this does NOT mean and are BOTH roots of , therefore it does NOT mean and are both factors of . ONLY if gives a remainder of 0 will be a factor of it. (or if divided through by )
Just because you can divide a polynomial by a linear expression, it does not strictly mean that linear expression has a root for the polynomial.
Practice long division with polynomial expressions to see what I mean.(Original post by RDKGames)
Just realised p(n) should be written as p(x) instead. Wouldn't make sense for a function of p to be expressed in terms of x, sorry. xD
But this is what I mean, therefore x=2 a real root if p(x)=0. If you plug in x=2, the first division will not be determined because you would be dividing by 0 on the denominator, thus x=2 cannot be a root.
Don't worry, I've done a lot of long divsion :P. Loads of questions and practice, mastered it. I just got a bit confused on a different question and realised this, hence why I asked. Thanks for the explanation though .
Posted from TSR Mobile 
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 25062016 20:29
(Original post by Chittesh14)
In all honesty, I knew where I was wrong and what it meant. I just wanted to see you how both would react to it and explain it to me !
Yeah , I realised that haha. Thanks for explaining it .Last edited by RDKGames; 25062016 at 20:34. 
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 25062016 20:33
(Original post by RDKGames)
Well lucky you I was bored enough to explain it detail, otherwise I would've left SeanFM's explanation for you :P 
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 25062016 20:33
(Original post by Chittesh14)
In all honesty, I knew where I was wrong and what it meant. I just wanted to see you how both would react to it and explain it to me !

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 25062016 20:35
OP please do me a favour and solve x^{2}+2x+2=0 , I'm so stuck on it!

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 25062016 20:41
(Original post by RDKGames)
OP please do me a favour and solve x^{2}+2x+2=0 , I'm so stuck on it!
Well, you can't complete the square either .
Well, the quadratic formula doesn't work either wtf lol, this equation has no solutions.
so there are no real roots.
Apparently, the solutions (roots) are imaginary lol so the graph does not cross the xaxis.
I tried WolframAlpha and it says 1 + i and 1  i.
I don't know what that means lol :/ sorry I can't help :'(.
Btw, what year are you in?  just wonderingLast edited by Chittesh14; 25062016 at 20:43. 
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 25062016 20:42
(Original post by SeanFM)
Riiight. Thanks for that ...
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Updated: July 21, 2016
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