#1

I came across this problem in an optional homework assignment and have little idea of what to do with it...

Could somebody explain to me some of the methods/properties they would use to solve for K to make it a true statement? I don't see how 3 root (a^9b^k) can possibly end up with a to the 3rd. Isn't 3 the

*square*of 9? So confused...

#2

3 is the square of 9, but consider the fact that K is the same in and out of the radical... that means that whatever happened in taking the cube root of the problem happened to make it come to (A^3)(B^k)

#3

It's optional, who cares.

#4

3 is the square of 9, but consider the fact that K is the same in and out of the radical... that means that whatever happened in taking the cube root of the problem happened to make it come to (A^3)(B^k)

Yea, that crossed my mind. I'm just having trouble thinking of a value that could do that o.o

I guess I'll just go through a list of numbers with strange properties like that... Or something.

It's optional, who cares.

I chose to do it in the first place because I need to understand the material well enough to perform well on the test. And I don't, as of this moment.

#5

Edit: maybe you weren't looking for that, but that's how powers/roots work.

#6

K= 1?

#7

Edit: maybe you weren't looking for that, but that's how powers/roots work.

Ah, thank you, that's EXACTLY it. I knew that I had something wrong there. Praise Maus24

EDIT:... So k=0 and 1?

#8

oh, and when you take a exponent to another exponent, you multiply. a.k.a..... (a^3)^3 = a^9

#9

I'd say that's a pretty radical expression!!

#10

to revise my previous post, K would equal 0, not one. (The cubic root of 1 is 1)

#11

to revise my previous post, K would equal 0, not one. (The cubic root of 1 is 1)

Correct sir is correct

#12

.....k can equal to any value and the equality would be correct....

#13

.....k can equal to any value and the equality would be correct....

wrong dude is wrong

#14

wrong dude is wrong

I read it wrong.... my bad....darn my ability to not be able to read properly

#15

Ah, another extremely confusing problem...

( 2^2001 )( 5^1950 )/ 4^27

I'm supposed to be finding the amount of digits in that number.

I'm obviously not supposed to actually calculate the number. I'm pretty sure it's going to end up being 10^something, because that's the only reliable way to get a specific number of zeros, atleast at our level. So... Do I multiply the bases and the exponents, or add, or what? Halp :S

( 2^2001 )( 5^1950 )/ 4^27

I'm supposed to be finding the amount of digits in that number.

I'm obviously not supposed to actually calculate the number. I'm pretty sure it's going to end up being 10^something, because that's the only reliable way to get a specific number of zeros, atleast at our level. So... Do I multiply the bases and the exponents, or add, or what? Halp :S

*Last edited by Firequacker at Mar 11, 2008,*