I was wondering if there was a way to simplify this problem instead of resorting to calculating the problem out by numerator and denominator. Thank you!
:::> [(8^2)(3^3)(2^4)]/(96^2)
I know this is a silly question but just to help brush up on my understanding of exponents would be a great help!
exponent problem
This topic has expert replies
-
- Senior | Next Rank: 100 Posts
- Posts: 55
- Joined: Tue Jul 27, 2010 8:37 pm
- Thanked: 11 times
I'm learning to like these problems! With multiplying/dividing exponents you should try to make the bases the same, so break down 96^2 into smaller numbers:
96^2 = (4 * 24)^2 = (2^2 * 4 * 6)^2
Maybe not the nicest way to do it but I think it's easy to look at 100 as 25*4 and then subtract 4 to get 24*4, and then you have the (2^2)^2 term so you can cancel that with 2^4. Now you have:
(8^2 * 3^3) / (4*6)^2
You need to get the 6 broken down to 2 and 3, so:
(8^2 * 3^3) / (4*2*3)^2
And then the 4*2 is 8 so you have:
8^2 * 3^3 / 8^2 * 3^2
The 8^2s cancel and you have 3^3 / 3^2 = 3.
It's harder to type than to write so maybe you can do it faster, but that's how I thought about it. Get the bases to be the same and you can start cancelling.
96^2 = (4 * 24)^2 = (2^2 * 4 * 6)^2
Maybe not the nicest way to do it but I think it's easy to look at 100 as 25*4 and then subtract 4 to get 24*4, and then you have the (2^2)^2 term so you can cancel that with 2^4. Now you have:
(8^2 * 3^3) / (4*6)^2
You need to get the 6 broken down to 2 and 3, so:
(8^2 * 3^3) / (4*2*3)^2
And then the 4*2 is 8 so you have:
8^2 * 3^3 / 8^2 * 3^2
The 8^2s cancel and you have 3^3 / 3^2 = 3.
It's harder to type than to write so maybe you can do it faster, but that's how I thought about it. Get the bases to be the same and you can start cancelling.