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Basic indices A level maths

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Original post by lizz-ie
I'm really sorry! :s-smilie:
By the way, was that AS or A2?


It's AS level
I gave you +ve rep for answering earlier, thanks for contributing
Original post by Philosophy&Econ

And for the one I just posted I got x=2 aswell(shall I post my working?)


Up to you, think that's right though, 3x=6
Reply 22
You need to use logs for this question:

9^2x=(1/3)^3

2xlog9=3log(1/3)
2x=(3log(1/3)) divided by (log9)
therefore, x=the above line, all divided by 2.

Hope this helps :smile:
Reply 23
Original post by AquaDrops
You need to use logs for this question:

9^2x=(1/3)^3

2xlog9=3log(1/3)
2x=(3log(1/3)) divided by (log9)
therefore, x=the above line, all divided by 2.

Hope this helps :smile:

You don't need logs. Read the whole thread.
Original post by notnek
You don't need logs. Read the whole thread.


This is because they have the same base number 3? So when it was whittled down to 3^(4x) = 3^(-3) to find x just look at the powers? Hence no need for logs?
Reply 25
Original post by notnek
You don't need logs. Read the whole thread.


While you may not neccessarily need them, the question is a whole lot easier if you do use them. You get the same answer anyway.
Reply 26
Original post by SubAtomic
This is because they have the same base number 3? So when it was whittled down to 3^(4x) = 3^(-3) to find x just look at the powers? Hence no need for logs?

Yes that's right - you can equate the powers.
Original post by AquaDrops
While you may not neccessarily need them, the question is a whole lot easier if you do use them. You get the same answer anyway.


False, logs makes it more complicated. Find the power and hey-presto no need for logs.

I thought logs was needed at first but can now see it is a waste of time.

I need to revise some things.
(edited 11 years ago)
Reply 28
Original post by AquaDrops
While you may not neccessarily need them, the question is a whole lot easier if you do use them. You get the same answer anyway.

In my opinion, the method is easier, faster and nicer if you don't use logs.
Original post by SubAtomic
Up to you, think that's right though, 3x=6


might aswell
Original post by Philosophy&Econ
might aswell


Yep, sorry for confusing you with logs as they are clearly not needed in these circumstances:colondollar: live and learn:biggrin:
(edited 11 years ago)
Original post by SubAtomic
Yep, sorry for confusing you with logs as they are clearly not needed in these circumstances:colondollar:


No problem :smile:
Thanks for the help anyway
Reply 32
Original post by notnek
In my opinion, the method is easier, faster and nicer if you don't use logs.


Hmm, I have to disagree with you on that one!

I just love logs you see haha :biggrin:
Reply 33
Original post by SubAtomic
False, logs makes it more complicated. Find the power and hey-presto no need for logs.

I thought logs was needed at first but can now see it is a waste of time.

I need to revise some things.


As notnek posted but cannot quote him for some reason:rolleyes:

xa+xba+b\displaystyle x^a + x^b \Longleftrightarrow a+b . Only if after finding the powers.

Was this what you posted notnek because it has vanished.


I don't think you can really say that my opinion is 'false' haha, but I suppose everyone is entitled to their own opinion :smile:

I love using logs, makes things easier for me.
Original post by AquaDrops
I don't think you can really say that my opinion is 'false' haha, but I suppose everyone is entitled to their own opinion :smile:

I love using logs, makes things easier for me.


Fair enough, I thought logs was needed at first but have been shown the light.

Saves a bit of space on the paper, saves some logs:tongue:
Reply 35
Original post by SubAtomic
False, logs makes it more complicated. Find the power and hey-presto no need for logs.

I thought logs was needed at first but can now see it is a waste of time.

I need to revise some things.


As notnek posted but cannot quote him for some reason:rolleyes:

xa+xba+b\displaystyle x^a + x^b \Longleftrightarrow a+b

I posted something similar to that but I realised that what I posted wasn't mathematically valid and I don't want to confuse anyone by using implications wrongly.
Original post by notnek
I posted something similar to that but I realised that what I posted wasn't mathematically valid and I don't want to confuse anyone by using implications wrongly.


Thought as much but didn't want to sound cheeky, will get rid of it:cool:
Reply 37
Original post by SubAtomic
Saves a bit of space on the paper, saves some logs:tongue:


I see what you did there.. haha!
Reply 38
Original post by SubAtomic
Thought as much but didn't want to sound cheeky, will get rid of it:cool:

Here's a better version of what I wrote:

If a is a positive real number, not equal to 1 then

ax=ay    x=y\displaystyle a^x = a^y \iff x=y
Original post by notnek
Here's a better version of what I wrote:

If a is a positive real number, not equal to 1 then

ax=ay    x=y\displaystyle a^x = a^y \iff x=y


Get ya:cool:

Going through a book at min with proofs and other various stuff, may call on your capable services in the near future on a few minor (depending how you look at it) details I maybe get but am unsure. New notation and sets, bit confusing sometimes as to what it is saying. All this stuff ⊈  \displaystyle \not\subseteq \ \exists \ \forall I know what these mean.

All the best
(edited 11 years ago)

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