Maths help needed!

Hey, doing some revision here and I just can't for the life of me work out how do this:

I needed to factorise this term by x:

xlog(a)
________ = 1
xlog(2b)

How do you factorise a fraction by a term which is on top and bottom?

Help would be appreciated, thanks

[MadGamer]

Im sure you divide it

[MetaWraith]

I would have said it doesnt factorise, since if you divide both top and bottom by x then x disappears from the equation altogther, leaving :

log(a) / log(2b) = 1

thus log(a) = log(2b)from which a=2b

Originally Posted by MetaWraith

I would have said it doesnt factorise, since if you divide both top and bottom by x then x disappears from the equation altogther, leaving :

log(a) / log(2b) = 1

thus log(a) = log(2b)from which a=2b

Yeah that's what I thought but unfortunately I was told find x?

I think the question is wrong.

[nffc]

It is. However you look at it x cancels.

[Graham M]

So x=0 ?

[Robert Atkins]

Originally Posted by Zeph

So x=0 ?

Nope, x can be any number but 0 provided that a=2b.

Are you sure you typed the equation correctly?

[Paul M]

x is anything you like, except 0 (you can't divide by zero).
__________________

Damn that Robert, beaten (just).

Thanks everyone

[bmxbandit]

Originally Posted by Paul M

x is anything you like, except 0 (you can't divide by zero).

it doesn't make sense to you computing folk, but i spent a good while in pure maths classes trying to figure out what 0/0 is in various cases...
__________________

if you rearrange such that log(a^x)=log((2b)^x), then theres no reason x cant be zero

apologies for pedantry!

[nffc]

Originally Posted by bmxbandit

if you rearrange such that log(a^x)=log((2b)^x), then theres no reason x cant be zero

apologies for pedantry!

Surely there is- as it's still re-arrangeable back to x log a = x log y, and that cancels. It's also still common to both sides as it is there as well.

[bmxbandit]

Originally Posted by nffc

Surely there is- as it's still re-arrangeable back to x log a = x log y, and that cancels. It's also still common to both sides as it is there as well.

...but the possiblity of it being zero is still there, you only 'cancel' to discard the trivial answer!

anyway, far too deep in. i'll shut up now

[Orior]

I've a similar problem (Is this off topic?)

£1 = 100 p
= (10p) * (10p)
= (£0.1) * (£0.1)
= £0.01
= 1p

So, £1 = 1 penny. What the!

[Paddy1]

Originally Posted by Orior

= (10p) * (10p)

eh no

= 10p * 10

[nffc]

Originally Posted by Paddy1

eh no

= 10p * 10

Exactly what I was going to say... you can't factorise as well as take the units to two quantities in the process.

Going from £1 = 100p to 10*10p is fine but multiplying 10p*10p gives 10p^2.