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A-level Maths - ln(x) - Natural Logs... HELP?

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Original post by Philip-flop
x


When dealing with e, use ln.

It's the same as going log to the base e. The reason this is so useful is lne = 1, making life much easier.
Original post by Philip-flop
Oh yay. At least I kind of got the question right. But how do I give my answer in terms of ln2\ln 2 ?? :frown:


express 8 as a power of 2 and go from there
Reply 22
Original post by Philip-flop
Oh yay. At least I kind of got the question right. But how do I give my answer in terms of ln2\ln 2 ?? :frown:


Watch this whole video https://www.youtube.com/watch?v=imKALH7eAZ4

then you'll find these questions more than doable...
Original post by Philip-flop
Oh yay. At least I kind of got the question right. But how do I give my answer in terms of ln2\ln 2 ?? :frown:


Is there a power relationship between 2 and 8 you can exploit?
Remembering that log(x^a)=alog(x)?
(edited 7 years ago)
Ok guys. With your help I think I managed to finally get the answer. As seen below...
Photo 03-09-2016, 11 12 48.jpg

Original post by Jasaron
When dealing with e, use ln.

It's the same as going log to the base e. The reason this is so useful is lne = 1, making life much easier.

Thank you. I decided to take your advice and replace log(e) with ln(e). Your help was very useful!

Original post by ValerieKR
Is there a power relationship between 2 and 8 you can exploit?
Remembering that log(x^a)=alog(x)?

Thanks for your help!! I didn't even think of replacing the 8 with a 2^3!! How did you know to do that?

Original post by Naruke
Watch this whole video https://www.youtube.com/watch?v=imKALH7eAZ4

then you'll find these questions more than doable...

I only watched half of that video before some things started making more sense! Thank you so much!! Going to watch the rest of the video today :smile:
Original post by Philip-flop

Thank you. I decided to take your advice and replace log(e) with ln(e). Your help was very useful!



Nope nope nope. You can't do that. log(e) and ln(e) are two completely different things, you cannot just switch between the two because they're not equal. log(e) is dealing with base 10 while ln(e) deals with base e which is why it goes away (as it's equal to 1). That would only be valid if you rewrote ln(e) as loge(e)log_e(e)
(edited 7 years ago)
Original post by RDKGames
Nope nope nope. You can't do that. log(e) and ln(e) are two completely different things, you cannot just switch between the two because they're not equal. log(e) is dealing with base 10 while ln(e) deals with base e which is why it goes away (as it's equal to 1). That would only be valid if you rewrote ln(e) as loge(e)log_e(e)


OMG. Oops I'm so stupid for having done that! I wrote the 4th line out completely wrong as it is :colondollar:
Ok. So I've scrapped my last answer and decided to start again. This time without taking logs to both sides! I'm probably still wrong though haha :frown: Sorry if I'm frustrating any of you guys :s-smilie:

Attachment not found
(edited 7 years ago)
Reply 28
Original post by RDKGames
Nope nope nope. You can't do that. log(e) and ln(e) are two completely different things, you cannot just switch between the two because they're not equal. log(e) is dealing with base 10 while ln(e) deals with base e which is why it goes away (as it's equal to 1). That would only be valid if you rewrote ln(e) as loge(e)log_e(e)


btw log=ln\log = \ln for almost every mathematician. log=log10\log = \log_{10} for engineers.
Reply 29
Original post by Philip-flop
Ok. So I've scrapped my last answer and decided to start again. This time without taking logs to both sides! I'm probably still wrong though haha :frown:



Yeah, that's correct
Original post by Philip-flop

Thanks for your help!! I didn't even think of replacing the 8 with a 2^3!! How did you know to do that?


Reducing log(a^x) to xlog(a) is a standard move with logs ^.^

I use the word logs to mean a logarithm with any base - in maths you'll take ln 99% of the time and in engineering 'log' will normally mean log base 10
(edited 7 years ago)
Original post by Zacken
Yeah, that's correct


Yayyyy. I finally got it. Still a bit confused but at least I'm getting there, right?!
Shame I don't know how to work this question out when I take logs to both sides :frown:
Original post by Philip-flop
Yayyyy. I finally got it. Still a bit confused but at least I'm getting there, right?!
Shame I don't know how to work this question out when I take logs to both sides :frown:


Your method still involves taking logs of both sides, it's just that those logs are the special case of logs called "natural logarithms" in this case. When you say take logs, that would mean the same thing regardless of the base.
Reply 33
Original post by Philip-flop
Yayyyy. I finally got it. Still a bit confused but at least I'm getting there, right?!
Shame I don't know how to work this question out when I take logs to both sides :frown:


You did take logs to both sides, just without knowing it. When moving from your first to your second line, you did it.

From: e3x=8e^{3x} = 8, you take loge\log_e to both sides to get logee3x=loge8\log_e e^{3x} = \log_e 8.

Using the power rule gets you 3xlogee=loge83x \log_e e = \log_e 8.

But you know that logaa=1\log_a a = 1 so logee=1\log_e e = 1 specifically.

So your equation becomes 3x=loge83x = \log_e 8 as you've written down. You did take logs to both sides, just without knowing it.
Original post by Zacken
You did take logs to both sides, just without knowing it. When moving from your first to your second line, you did it.

From: e3x=8e^{3x} = 8, you take loge\log_e to both sides to get logee3x=loge8\log_e e^{3x} = \log_e 8.

Using the power rule gets you 3xlogee=loge83x \log_e e = \log_e 8.

But you know that logaa=1\log_a a = 1 so logee=1\log_e e = 1 specifically.

So your equation becomes 3x=loge83x = \log_e 8 as you've written down. You did take logs to both sides, just without knowing it.

Oh yeah!! Thank you so much for clearing that up for me!! That actually makes perfect sense! You made me realise that I did take logs to both sides its just the LHS was "hidden" in a way.

Original post by RDKGames
Your method still involves taking logs of both sides, it's just that those logs are the special case of logs called "natural logarithms" in this case. When you say take logs, that would mean the same thing regardless of the base.

Yes I'm still trying to get used to the fact that... loge(x)\log_e (x) is the same as ln(x)\ln (x) :colondollar:
Reply 35
Original post by Philip-flop
Oh yeah!! Thank you so much for clearing that up for me!! That actually makes perfect sense! You made me realise that I did take logs to both sides its just the LHS was "hidden" in a way.


No worries.

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