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# Laws of Exponents

Directions: Use the integers 1 through 20 at most one time each, and fill in the blanks to create equivalent expressions. ### Hint

What types of numbers are easy to arrange into products?

Source: Shaun Errichiello

## Minimize Slope

Directions: Given the point (3,5), use digits 1-9, at most one time, to find a …

1. This problem was fantastic! My awesomely skilled sixth graders found seven more solutions to this problem! I wonder if we found them all? Thank you!

2. I see that some students have found more solutions. I thought I might share a solution found by one of my 8th graders.

(2^2)^5 = (2^6)^11/(2^7)^8 = 2^3 x 2^7 = 2^20/2^10

• We found a mistake ( that is after I double checked. ugh!) change 2^3 x 2^7 to 2^1 x 2^9

3. I think I found a ton that all equal 32. First I rewrote it:

(2^a)^b = (2^c)/(2^d) = 2^g * 2^h = 2^j / 2^k

Then {a,b,c,d,e,f,g,h,j,k} can be

1,5,4,17,7,9,2,3,11,6
1,5,4,17,7,9,2,3,13,8
1,5,4,17,7,9,2,3,15,10
1,5,4,17,7,9,2,3,16,11
1,5,4,17,7,9,2,3,18,13
1,5,4,17,7,9,2,3,19,14
1,5,4,17,7,9,2,3,20,15
1,5,7,11,4,18,2,3,15,10
1,5,7,11,4,18,2,3,17,12
1,5,7,11,4,18,2,3,19,14
1,5,7,11,4,18,2,3,20,15

or any permutation of any of these that switches the numbers in the a,b and/or c,d and/or e,f and/or g,h pairs.

I suspect there are a lot of other solutions too!

4. I got a different answer than the one provided. I got:
(2^2)^3 = (2^15)^6/(2^7)^12= 2^1 x 2^5 = 2^10/2^4

Did anyone else get this answer? Is this correct?