Maths help needed!

[Macca371]

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

[Macca371]

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]

So x=0 ? :)

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

Are you sure you typed the equation correctly?

[Paul]

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

Damn that Robert, beaten (just). :)

[Macca371]

Thanks everyone :tu:

[bmxbandit]

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 :D

apologies for pedantry!

[nffc]

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

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]

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 :erm: :D

[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]

= (10p) * (10p)

eh no :)

= 10p * 10

[nffc]

= (10p) * (10p)

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.

[Paul]

10p^2.Is that Square Pennies :erm:

Okay, I'll get me coat ...... :walk:

[nffc]

10p^2.Is that Square Pennies :erm:

Okay, I'll get me coat ...... :walk:
not funny. And yes I have got out of the unfunny side of bed :rolleyes:

[ScaredWebWarrior]

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 :)

It's too long ago for me to remember this stuff, lol, but I'm sure the properties of logarithms must play a part:




logb(xy) = logbx + logby.
logb(x/y) = logbx - logby.
logb(xn) = n logbx.
logbx = logax / logab.
See also: http://oakroadsystems.com/math/loglaws.htm (http://oakroadsystems.com/math/loglaws.htm)
or http://tutorial.math.lamar.edu/AllBrowsers/1314/SolveLogEqns.asp

[Theodoric]

It's too long ago for me to remember this stuff, lol, but I'm sure the properties of logarithms must play a part:




logb(xy) = logbx + logby.
logb(x/y) = logbx - logby.
logb(xn) = n logbx.
logbx = logax / logab.
See also: http://oakroadsystems.com/math/loglaws.htm (http://oakroadsystems.com/math/loglaws.htm)
or http://tutorial.math.lamar.edu/AllBrowsers/1314/SolveLogEqns.asp
I dunno. As a number of people have said, and I agree with them, there is no way that the x's won't cancel out, in which case you must end up with a = 2b, that is a and b both zero.

However, and my memory is rapidly failing me here, if a and b are complex numbers then their logs can have multiple values (possibly :)).

[bmxbandit]

However, and my memory is rapidly failing me here, if a and b are complex numbers then their logs can have multiple values (possibly :)).
log[x+iy] = log[mod(x+iy)] + i [2*pi*n*arg(x+iy)]

where n is a natural number, or 0. can't remember if this is just natural log, its a few years since i did complex maths!
__________________

makes no difference to the answer though :(

damn, that took a while to work out!

[Theodoric]

log[x+iy] = log[mod(x+iy)] + i [2*pi*n*arg(x+iy)]

where n is a natural number, or 0. can't remember if this is just natural log, its a few years since i did complex maths!
__________________

makes no difference to the answer though :(

damn, that took a while to work out!
Can I take that that "n" means that I was vaguely correct? :)

[bmxbandit]

Can I take that that "n" means that I was vaguely correct? :)
yes indeedy :D