Given the integer A, can

log (A log A) be simplified somehow? Assume both logs have a base of 2.

4/7/2007 12:45:20 PM

do you know A?

log (A * log A) =

log A + log (log A)

i don't think it can be simplified beyond that, unless you either know A, or the know what the original expression equals.

or you can do:

log (A * log A) =

log [log (A^A)]

again that's it.

4/7/2007 1:11:21 PM

I'm just horsing around here but...

Log (A Log A)=Log(LogA^A) OEPII1

2^X=Log(A^A) You said they were base 2

A^A=2^(2^X)... yea I got nothing... I agree with OEPII1 that's a simplified as it's gonna get.

4/9/2007 11:56:12 AM

do it this way and they'll take off points from other parts of the assignment,

y= log(A log(A))

solving for A gives,

A^2 = y / log^2

so take the squareroot,

A = sqrt( y/log^2)

there's what A is

4/9/2007 11:07:24 PM

^take his advice

he's the mathman

4/9/2007 11:14:20 PM

^^ i was gonna do that, but he didn't say we had to solve for A, because well, he didn't give an equation, rather just an expression.

he explicity asked for simplifying an expression he gave, not for solving the equation he didn't give. who knows what he wants...?

but yeah, if he wants to solve for A, then that's what he should do, BUT he has has to know what the original expression equals (y, which you introduced). without that, it is impossible.

Quote :

"i don't think it can be simplified beyond that, unless you either know A, or the know what the original expression equals."

4/10/2007 8:35:54 AM

i just assumed mathman had done it correctly, but i just looked at it now, and i am like, what the hell has he done???

Quote :

"do it this way and they'll take off points from other parts of the assignment, y= log(A log(A)) solving for A gives, A^2 = y / log^2 --> how the hell do you get that??? and WORSE, what is "log^2" ????? so take the squareroot, A = sqrt( y/log^2) there's what A is "

omg, is this for real????

unbelievable. how can you have a username like that, and then write that utter nonsense??? i just hope that was a joke on his part.

if we assume this:

y= log(A log A)

then this can be done:

y = log [log (A^A)]

2^y = log (A^A)

2^(2^y) = A^A

which is exactly what wizzkidd got.

4/10/2007 1:36:20 PM

Quote :

"do it this way and they'll take off points from other parts of the assignment,"

4/10/2007 7:33:08 PM

^^hahahaa the funniest part is that you didn't catch the joke the first time and even agreed with it. I just assumed you knew what was going on and were playing along.

A++++ would read again!

4/10/2007 8:13:08 PM

Quote :

"hahahaa the funniest part is that you didn't catch the joke the first time and even agreed with it."

i didn't even look at it the first time beyond the first line, as i said before. i looked at it hours later.

btw, i still don't get what this means, which i guess is the joke:

Quote :

"do it this way and they'll take off points from other parts of the assignment"

4/11/2007 7:05:44 AM

yeah, I didn't really get that either. But that math had to be a joke.

[Edited on April 11, 2007 at 8:29 AM. Reason : .]

4/11/2007 8:28:22 AM

I mean if I see something that wrong on some assignment I will make it my mission to take off as
many points as possible, no mercy, cause that is just plain stupid and should be punished.

4/11/2007 8:09:50 PM

ah ok.

but that's a common mistake that a lot of non-science/eng/math students make if they have never had a good understanding of the topic of function. which is sad, because it is such an easy topic to teach AND to understand, and fun too. being a math teacher, i see it all the time.

happens with trig and log functions the most.

sin x = 0.5

x = 0.5/sin

4/11/2007 8:59:03 PM

ya, its pretty common, and in precalculus or algebra its sort of understandable, but when I see it in a calculus course it makes me

4/11/2007 11:15:52 PM