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good kid, maad city - a small-town kid's A-level journey

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Reply 540
Original post by Zacken
This needs updating. :talkhand:


Original post by Student403
Yes!! Awfully quiet couple of days


I have a pretty big update for tonight :biggrin:
Original post by aymanzayedmannan
I have a pretty big update for tonight :biggrin:


Ooh la la. I'm intrigued. :biggrin:
Um, didn't feel like making a thread for this, but why is

Unparseable latex formula:

e^l^n^x = ln(e^x) ?



I understand the later that ln(e^x) = x

xln(e)=x1=xx * ln(e) = x * 1 = x

(I hope it's fine to ask here):s-smilie:
(edited 8 years ago)
Original post by SaadKaleem
Um, didn't feel like making a thread for this, but why is

Unparseable latex formula:

e^l^n^x = ln(e^x) ?



I understand the later that ln(e^x) = x

xln(e)=x1=xx * ln(e) = x * 1 = x

(I hope it's fine to ask here):s-smilie:


exe^x and lnx\ln x are inverse operations. So composing them together gets you xx.

i.e: f(x)f1(x)=xf(x) \circ f^{-1}(x) = x.

Here we have f(x)=exf1(x)=lnxf(x) = e^x \Rightarrow f^{-1}(x) = \ln x or f(x)=lnxf1(x)=exf(x) = \ln x \Rightarrow f^{-1}(x) = e^x.

It's the same reason why x2=x\sqrt{x^2} =x, because x\sqrt{x} and x2x^2 are inverse functions.
Reply 544
Original post by SaadKaleem
Um, didn't feel like making a thread for this, but why is

Unparseable latex formula:

e^l^n^x = ln(e^x) ?



I understand the later that ln(e^x) = x

xln(e)=x1=xx * ln(e) = x * 1 = x

(I hope it's fine to ask here):s-smilie:



It's alright man! Zacken is on this thread so no worries. I was going to say they're inverse functions, that's why - but I knew he'd come up with a much more concise explanation.
Original post by Zacken
exe^x and lnx\ln x are inverse operations. So composing them together gets you xx.

i.e: f(x)f1(x)=xf(x) \circ f^{-1}(x) = x.

Here we have f(x)=exf1(x)=lnxf(x) = e^x \Rightarrow f^{-1}(x) = \ln x or f(x)=lnxf1(x)=exf(x) = \ln x \Rightarrow f^{-1}(x) = e^x.

It's the same reason why x2=x\sqrt{x^2} =x, because x\sqrt{x} and x2x^2 are inverse functions.


Your explanations of stuff like this are just brilliant
Reply 546
Original post by Zacken
It's the same reason why x2=x\sqrt{x^2} =x, because x\sqrt{x} and x2x^2 are inverse functions.


However, this only applies on a certain domain, right?
Original post by Zacken
exe^x and lnx\ln x are inverse operations. So composing them together gets you xx.

i.e: f(x)f1(x)=xf(x) \circ f^{-1}(x) = x.

Here we have f(x)=exf1(x)=lnxf(x) = e^x \Rightarrow f^{-1}(x) = \ln x or f(x)=lnxf1(x)=exf(x) = \ln x \Rightarrow f^{-1}(x) = e^x.

It's the same reason why x2=x\sqrt{x^2} =x, because x\sqrt{x} and x2x^2 are inverse functions.


Excellent, Thank you!


Original post by Student403
Your explanations of stuff like this are just brilliant

I concur :wink:
(edited 8 years ago)
Original post by aymanzayedmannan
However, this only applies on a certain domain, right?


If I was on the maths forum I'd have put in a bracket to specify:

(blah blah, yes, this is only true if we restrict the domain of xx2x \mapsto x^2 to be an injective function i.e, we need x0x\geq 0).

But I figured I needn't on here. :tongue:
Original post by aymanzayedmannan
...


BTW: Big update? :tongue:

Original post by Student403
Your explanations of stuff like this are just brilliant


Much appreciated. :biggrin:
Reply 550
Original post by Zacken
If I was on the maths forum I'd have put in a bracket to specify:

(blah blah, yes, this is only true if we restrict the domain of xx2x \mapsto x^2 to be an injective function i.e, we need x0x\geq 0).

But I figured I needn't on here. :tongue:


I was only teasing you lad :tongue: I like how you can change gears from Maths nerd Zacken to bantz Zacken so easily.


Original post by Zacken
BTW: Big update? :tongue:


I'll post it first thing tomorrow morning! I did a lot of conic sections tonight and I feel like I don't suck AS bad anymore... :laugh:

When are you doing your mocks? I tried the S3 paper and the first 4 questions were very nice, but I couldn't proceed past it because I'd didn't cover contingency tables or gof for continuous distributions. You'll love a few of those questions though!
Original post by aymanzayedmannan
I was only teasing you lad :tongue: I like how you can change gears from Maths nerd Zacken to bantz Zacken so easily.




I'll post it first thing tomorrow morning! I did a lot of conic sections tonight and I feel like I don't suck AS bad anymore... :laugh:

When are you doing your mocks? I tried the S3 paper and the first 4 questions were very nice, but I couldn't proceed past it because I'd didn't cover contingency tables or gof for continuous distributions. You'll love a few of those questions though!


His post yesterday or the day before killed me.

"HA GOT YOU"

:toofunny:
Reply 552
Original post by Student403
His post yesterday or the day before killed me.

"HA GOT YOU"

:toofunny:


Zacken420 XD that entire thread was on drugs...
Original post by aymanzayedmannan
Zacken420 XD that entire thread was on drugs...


I honestly had never seen that part of him come alive before :rofl:
Reply 554
Original post by Student403
I honestly had never seen that part of him come alive before :rofl:


He also admitted that he has a vagina, let's not forget :rofl: He's definitely been hitting some good stuff after results day. 🌚
Original post by aymanzayedmannan
He also admitted that he has a vagina, let's not forget :rofl: He's definitely been hitting some good stuff after results day. 🌚


@Zacken420 hook us up pal will ya? :laugh:
Original post by aymanzayedmannan
He also admitted that he has a vagina, let's not forget :rofl:


say what?
Original post by tinkerbella~
say what?


He implied that he menstruates
Original post by Student403
He implied that he menstruates


Oh, that explains the moodiness every month then
Original post by tinkerbella~
Oh, that explains the moodiness every month then


Oh, yeah. I'm the moody one, yeah? :rolleyes:

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