very hard question

• Nov 19th 2009, 06:51 AM
andyboy179
very hard question
hi this is my question:

(2 x 3^7) x ( 3 x 2^5) = 2^n x 3^m find n and m

i am VERY confused on what to do, can anyone help please?!?
• Nov 19th 2009, 07:50 AM
BabyMilo
put it this way:
(2 x 2^5) x ( 3 x 3^7) = 2^n x 3^m
can you do it now?
• Nov 19th 2009, 07:53 AM
andyboy179
Quote:

Originally Posted by BabyMilo
put it this way:
(2 x 2^5) x ( 3 x 3^7) = 2^n x 3^m
can you do it now?

how did u get those numbers? some have changed
• Nov 19th 2009, 08:00 AM
BabyMilo

read it like this (2 x 3^7) x ( 3 x 2^5) = 2 x 3^7 x 3 x 2^5 right?
then rearrange a bit 2 x 3^7 x 3 x 2^5 = 2 x 2^5 x 3 x 3^7

you can add the brackets in if you want then.
• Nov 19th 2009, 08:01 AM
andyboy179
sorry i understand now, i was being dumb(Sleepy). is that all i would write?
• Nov 19th 2009, 08:04 AM
andyboy179
Quote:

Originally Posted by BabyMilo

read it like this (2 x 3^7) x ( 3 x 2^5) = 2 x 3^7 x 3 x 2^5 right?
then rearrange a bit 2 x 3^7 x 3 x 2^5 = 2 x 2^5 x 3 x 3^7

you can add the brackets in if you want then.

what would the answer be? or would i just put (2 x 2^5) x ( 3 x 3^7)
• Nov 19th 2009, 08:11 AM
BabyMilo
2 x 2^5 = 2^6
3 x 3^7 = 3^8
so n=6
and m=8

think of it as 2 x (2x2x2x2x2) and 3 x (3x3x3x3x3x3x3)
• Nov 19th 2009, 08:19 AM
andyboy179
Quote:

Originally Posted by BabyMilo
2 x 2^5 = 2^6
3 x 3^7 = 3^8
so n=6
and m=8

think of it as 2 x (2x2x2x2x2) and 3 x (3x3x3x3x3x3x3)

i understand now, thats alot(Clapping)