Year 12s: what will be the first way you research your future options? >>

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

Ok so last night I had trouble trying to figure out how this makes sense. I can't get my head round why this is...
Natural Logs.png

Can someone please explain? :frown:

Scroll to see replies

Reply 1

RDKGames

Study Forum Helper

Original post by Philip-flop

Ok so last night I had trouble trying to figure out how this makes sense. I can't get my head round why this is...
Natural Logs.png

Can someone please explain? :frown:

You should've covered this in C2. It's one of the logarithm rules. Or are you asking for the proof?

(edited 7 years ago)

Reply 2

Zacken

Original post by Philip-flop

Ok so last night I had trouble trying to figure out how this makes sense. I can't get my head round why this is...
Natural Logs.png

Can someone please explain? :frown:

It's the rule that

a \ln x = \ln x^a

.

Let

y = \ln x

then

x = e^y

.

So

x^a = (e^y)^a = e^{ay}

.

Take the logarithm of both sides

\ln x^a = \ln e^{ay} = ay

.

So

\ln x^a = a\ln x

Reply 3

ValerieKR

Original post by Philip-flop

Ok so last night I had trouble trying to figure out how this makes sense. I can't get my head round why this is...
Natural Logs.png

Can someone please explain? :frown:

You know
ln(x) + ln(y) =ln(xy)
let y=x
ln(x) +ln (x) = ln(x^2)
2ln(x) = ln(x^2)
You can extend the rule (forwards and backwards) to end up with any rational power and it still works (for irrationals you need Zacken's method)

(edited 7 years ago)

Reply 4

Philip-flop

Original post by RDKGames

You should've covered this in C2. It's one of the logarithm rules. Or are you asking for the proof?

Oh my god. Yeah of course!! I completely forgot! My Maths brain still hasn't switched on since the summer!

Original post by Zacken

It's the rule that

a \ln x = \ln x^a

.

Let

y = \ln x

then

x = e^y

.

So

x^a = (e^y)^a = e^{ay}

.

Take the logarithm of both sides

\ln x^a = \ln e^{ay} = ay

.

So

\ln x^a = a\ln x

Thank you Zacken!!
Can always rely on you for amazing explanations :smile:

Original post by ValerieKR

You know
ln(x) + ln(y) =ln(xy)
let y=x
ln(x) +ln (x) = ln(x^2)
2ln(x) = ln(x^2)
You can extend the rule to any power and it still works

Thank you so much!! Can't believe I forgot the "Powers of Logs" rule :colondollar:

Reply 5

Paulxo

Hi mate - I'll explain this stuff to you next week if you like ;-)

Reply 6

Philip-flop

Original post by Paulxo

Hi mate - I'll explain this stuff to you next week if you like ;-)

Don't you worry, I have a looooooong list of questions I need to ask you :P

Reply 7

Zacken

Original post by Philip-flop

Oh my god. Yeah of course!! I completely forgot! My Maths brain still hasn't switched on since the summer!

Thank you Zacken!!
Can always rely on you for amazing explanations :smile:

Thank you so much!! Can't believe I forgot the "Powers of Logs" rule :colondollar:

No worries

Reply 8

Philip-flop

Ok I seem to be slipping up on Natural Logs again FFS. Why can't I do part a and c? :frown:

For part c I end up with

x=\frac{\ln(0.5)-1}{2}

(edited 7 years ago)

Reply 9

RDKGames

Study Forum Helper

Original post by Philip-flop

Ok I seem to be slipping up on Natural Logs again FFS. Why can't I do part a and c? :frown:

For part c I end up with

x=\frac{\ln(0.5)-1}{2}

8=2^3

0.5=\frac{1}{2}=2^{-1}

Wait, how can you not do A and C when you've done B then...? I'm assuming you've done B since you're not asking for help on it...

(edited 7 years ago)

Reply 10

ValerieKR

Original post by Philip-flop

For part c I end up with

x=\frac{\ln(0.5)-1}{2}

^.-

Reply 11

Philip-flop

Original post by RDKGames

8=2^3

0.5=\frac{1}{2}=2^{-1}

Wait, how can you not do A and C when you've done B then...? I'm assuming you've done B since you're not asking for help on it...

I think I worked out part b as a fluke :frown:

Here's my answer for part b...
Photo 02-09-2016, 22 21 01.jpg

Am I rubbish at Logs? :frown:

Reply 12

RDKGames

Study Forum Helper

Original post by Philip-flop

I think I worked out part b as a fluke :frown:

Here's my answer for part b...

Am I rubbish at Logs? :frown:

It's technically correct but doesn't really flow well towards the end without converting the 4 into a power of 2.

Reply 13

ValerieKR

Original post by Philip-flop

Ok I seem to be slipping up on Natural Logs again FFS. Why can't I do part a and c? :frown:

For part c I end up with

x=\frac{\ln(0.5)-1}{2}

You're correct on part c - Remember that aln(x)= ln(x^a) even when a is negative

start part a by taking logs of both sides

Reply 14

Paulxo

On b, how did you turn ln 4/2 into ln 2?

Reply 15

Philip-flop

Original post by Paulxo

On b, how did you turn ln 4/2 into ln 2?

Because

\frac{1}{2}\times4

? I don't know :frown:

Reply 16

Philip-flop

Original post by ValerieKR

You're correct on part c - Remember that aln(x)= ln(x^a) even when a is negative

start part a by taking logs of both sides

Ok I've gone wrong with part a as well :frown:

Unparseable latex formula:

\loge^3x=\log8

Unparseable latex formula:

3x\loge=log8

Unparseable latex formula:

x=\frac{\log8}{3\loge}

Reply 17

Philip-flop

Great. I don't know how to even write e^3x in latex. Here is my answer below...
Photo 02-09-2016, 23 27 28.jpg

I am aware that I am completely wrong with my answer so try to look past my stupidity and help me if you can :frown:

(edited 7 years ago)

Reply 18

Naruke

Original post by Philip-flop

Great. I don't know how to even write e^3x in latex. Here is my answer below...
Photo 02-09-2016, 23 27 28.jpg

I am aware that I am completely wrong with my answer so try to look past my stupidity and help me if you can :frown:

This is correct. However, if this is Exercise 3B, Question 2A you need to give your answer in terms of

\ln 2

.

Don't call yourself stupid just because you can't solve a math problem. Everyone who is assisting you in this thread has been doing maths way longer than you. So, they'll obviously be a little bit sharper at it...