Simplifying exponents

G- Unit30 wrote:Hi, I had trouble with this Question from a GMATprep practice exam:

2^5 + 2^5 + 3^5 +3^5 + 3^5 =?

First, notice that K + K = 2K
Using the same logic, 2^5 + 2^5 = 2(2^5) = (2^1)(2^5) = 2^6

Similarly, notice that M + M + M = 3M
Using the same logic, 3^5 + 3^5 + 3^5 = 3(3^5) = (3^1)(3^5) = 3^6

So, 2^5 + 2^5 + 3^5 + 3^5 + 3^5 = 2^6 + 3^6

Aside: For some reason, I remember the question differently.
I think it's (2^5 + 2^5)(3^5 + 3^5 + 3^5) = ?
If that's the correct question, then we can evaluate it as follows:
2^5 + 2^5)(3^5 +3^5 + 3^5) = (2^6)(3^6)
= 6^6 [we're using the rule that says (x^n)(y^n) = (xy)^n]

Cheers,
Brent

DCS80 wrote:First, notice that K + K = 2K
Using the same logic, 2^5 + 2^5 = 2(2^5) = (2^1)(2^5) = 2^6

Similarly, notice that M + M + M = 3M
Using the same logic, 3^5 + 3^5 + 3^5 = 3(3^5) = (3^1)(3^5) = 3^6


can you explain how the first and second both equate to ^6? (given two factors versus three?)

You bet.

1 thing + 1 thing = 2things
2^5 + 2^5 = 2(2^5)
At this point, we can replace 2 with 2^1 to get: (2^1)(2^5)
Apply the product law to get: 2^6

Aside: Product Law: (x^a)(x^b) = x^(a+b)

Similarly, 1 thing + 1 thing + 1 thing = 3things
So, 3^5 + 3^5 + 3^5 = 3(3^5)
Replace 3 with 3^1 to get: (3^1)(3^5)
Apply the product law to get: 3^6

Cheers,
Brent