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C3 The exponential function question

IMG_20130711_185329-1.jpgIMG_20130711_185758-1.jpg
I need help on q20 and here's what I've done:

ln(x+6)=ln(x-6)^2

x+6=(x-6)^2

X^2 -13x + 30 = 0

(x-3)(x-10)=0

x=3 or x=10

So in the textbook answer which is attached why did they exclude the solution x=3?

Thanks :biggrin:

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Reply 1
Avatar for halpme
halpme
7
Because you can't take the natural log of a negative number and if x=3x=3 then 2ln⁡(x−6)2\ln(x-6) would be 2ln⁡(−3)2\ln(-3), which can't happen.
(edited 10 years ago)
Reply 2
Avatar for krisshP
krisshP
OP
14
Original post by halpme
Because you can't take the natural log of a negative number and if x=3x=3 then 2ln⁡(x−6)2\ln(x-6) would be 2ln⁡(−3)2\ln(-3), which can't happen.


I understand it now, so thank you:smile:
Reply 3
Avatar for krisshP
krisshP
OP
14
Original post by halpme
Because you can't take the natural log of a negative number and if x=3x=3 then 2ln⁡(x−6)2\ln(x-6) would be 2ln⁡(−3)2\ln(-3), which can't happen.

Hey can you help me in q22 please? I'm unsure on how to start it off and would like a small hint.

Thanks
Reply 4
Avatar for joostan
joostan
13
Original post by krisshP
Hey can you help me in q22 please? I'm unsure on how to start it off and would like a small hint.

Thanks


Perhaps consider and rearrange:y=log⁡a(x)y = \log_a(x) into the desried form. :smile:
Reply 5
Avatar for krisshP
krisshP
OP
14
Original post by joostan
Perhaps consider and rearrange:y=log⁡a(x)y = \log_a(x) into the desried form. :smile:


Wow that turned out simpler than I thought.:smile:
Attachment not found

Can you help me on that ^ q please. There's no textbook answer and here's what I've done.

Due to the AP
y/x =r [1]

z/y=r [2]
Substitute [1] into [2]
y/x = z/y

y^2=zx

2logy=logz + logx [5]

Due to the GP
logz -logy=d [3]

logy - logx=d [4]

Substitute [3] into [4]

logz -logy =logy -logx

2logy=logz + logx [6]

[5] and [6] are the same and match up, so there is no contradiction. QED

Is that an okay way to prove it? Also did I use QED right there? I think it means that a statement is fully proved now, right? Sorry for very long method.

Thanks for the help :biggrin:
Reply 6
Avatar for joostan
joostan
13
Original post by krisshP
Wow that turned out simpler than I thought.:smile:
Attachment not found

Can you help me on that ^ q please. There's no textbook answer and here's what I've done.

Due to the AP
y/x =r [1]

z/y=r [2]
Substitute [1] into [2]
y/x = z/y

y^2=zx

2logy=logz + logx [5]

Due to the GP
logz -logy=d [3]

logy - logx=d [4]

Substitute [3] into [4]

logz -logy =logy -logx

2logy=logz + logx [6]

[5] and [6] are the same and match up, so there is no contradiction. QED

Is that an okay way to prove it? Also did I use QED right there? I think it means that a statement is fully proved now, right? Sorry for very long method.

Thanks for the help :biggrin:

QED is often considered pretentious for a minor proof though it's use is correct.
Btw the attachment doesn't work but the working looks OK. :smile:
Reply 7
Avatar for krisshP
krisshP
OP
14
Original post by joostan
QED is often considered pretentious for a minor proof though it's use is correct.
Btw the attachment doesn't work but the working looks OK. :smile:


IMG_20130712_093628-1.jpg
Reply 8
Avatar for joostan
joostan
13
Original post by krisshP
IMG_20130712_093628-1.jpg


I'd have started off with:
GP:
x, y=rx, z=r2xx, \ y=rx , \ z=r^2x
Then:
Unparseable latex formula:

\log(x) , \ \log(r) + \log(y) \ \2\log(r) + log(z)[br]\Rightarrow \mathrm{AP \ with \ common\ ratio} \log(r)


Note also that permuting x,y and z has no effect so arbitrary selection of x, y, z has no effect on proof. :smile:
Reply 9
Avatar for krisshP
krisshP
OP
14
Original post by joostan
I'd have started off with:
GP:
x, y=rx, z=r2xx, \ y=rx , \ z=r^2x
Then:
Unparseable latex formula:

\log(x) , \ \log(r) + \log(y) \ \2\log(r) + log(z)[br]\Rightarrow \mathrm{AP \ with \ common\ ratio} \log(r)


Note also that permuting x,y and z has no effect so arbitrary selection of x, y, z has no effect on proof. :smile:


Everything after the bold "then" part above I can't understand.
Reply 10
Avatar for krisshP
krisshP
OP
14
How did you introduce r into the AP :eek: , I expected that you would involve d.
Original post by krisshP
Everything after the bold "then" part above I can't understand.


Thinking out loud, here are some hints from a quick scan:

Geometric has a common ratio so y/x = z/y
Take log

Arithmetic has a common difference so log y - log x = log z - log y

Can you put the 2 together?
Reply 12
Avatar for krisshP
krisshP
OP
14
Original post by m4ths/maths247
Thinking out loud, here are some hints from a quick scan:

Geometric has a common ratio so y/x = z/y
Take log

Arithmetic has a common difference so log y - log x = log z - log y

Can you put the 2 together?


What do you mean? I've now established log y - log x = log z - log y and

r= log y - log x
r= log z - log y
Original post by krisshP
What do you mean? I've now established log y - log x = log z - log y and

r= log y - log x
r= log z - log y


I really don't think the question needs much more than what I have put.

y/x = z/y for a GP as that is the ratio
Take logs to give log y/x = log z/y
Log Laws gives log y - log x = log z - log y

Move onto the AP
If its an AP and the first 3 terms are log x, log y and log z they have a common difference therefore log y - log x = log z - log y

You have already shown this from the GP therefore any conclusion from here in would be fine.

There is no requirement to introduce a,r,d etc. You are simply using the fact of common ratio and common difference for GP and AP respectively.
Reply 14
Avatar for krisshP
krisshP
OP
14
Original post by m4ths/maths247
I really don't think the question needs much more than what I have put.

y/x = z/y for a GP as that is the ratio
Take logs to give log y/x = log z/y
Log Laws gives log y - log x = log z - log y

Move onto the AP
If its an AP and the first 3 terms are log x, log y and log z they have a common difference therefore log y - log x = log z - log y

You have already shown this from the GP therefore any conclusion from here in would be fine.

There is no requirement to introduce a,r,d etc. You are simply using the fact of common ratio and common difference for GP and AP respectively.
IMG_20130713_102130-1.jpg
Okay, I understand now, thanks a lot:smile:

Can you please help me in q 7? See the attachment. I feel clueless on how to even start it off :frown: please help me out


THANKS :biggrin:
Original post by krisshP
IMG_20130713_102130-1.jpg
Okay, I understand now, thanks a lot:smile:

Can you please help me in q 7? See the attachment. I feel clueless on how to even start it off :frown: please help me out


THANKS :biggrin:


Have you done numerical methods yet?
You can write a function, say f(x), set it to 0 and evaluate f(1) and f(2) to show a change of sign in a continuous interval.
Reply 16
Avatar for krisshP
krisshP
OP
14
Original post by m4ths/maths247
Have you done numerical methods yet?
You can write a function, say f(x), set it to 0 and evaluate f(1) and f(2) to show a change of sign in a continuous interval.


Nope. BTW isn't that new method a bit like trial and error/improvement? What do you mean by a change of sign?
Reply 17
Avatar for MissAF01
MissAF01
Original post by krisshP
Nope. BTW isn't that new method a bit like trial and error/improvement? What do you mean by a change of sign?

a change of sign between f(1) and f(2) indicates that a root lies between x=1 and x=2.
Original post by krisshP
Nope. BTW isn't that new method a bit like trial and error/improvement? What do you mean by a change of sign?


I think it would be better to learn the topic first.
If you are yet to learn it you could consider the graphs of y = x and y ln(x+5) ot even e^x = x+5 to get some sense of a root but an understanding of numerical methods (learned from the beginning upwards) would mean the question is fairly straight forward.

It would seem the book you are using expects some understanding of the topic prior to it?
Reply 19
Avatar for krisshP
krisshP
OP
14
Original post by m4ths/maths247
I think it would be better to learn the topic first.
If you are yet to learn it you could consider the graphs of y = x and y ln(x+5) ot even e^x = x+5 to get some sense of a root but an understanding of numerical methods (learned from the beginning upwards) would mean the question is fairly straight forward.

It would seem the book you are using expects some understanding of the topic prior to it?


The book is this:
http://www.amazon.co.uk/Pure-Mathematics-C3-David-Rayner/dp/1902214463/ref=sr_1_1?ie=UTF8&qid=1373713049&sr=8-1&keywords=C3+C4+DAVID

It does give a few questions which tend to be slightly more challenging. I guess for this question they try to combine the topic numerical methods into it as a surprise.

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