2^5+2^5+3^5+3^5+3^5 is equal to:

a) 5^6
b) 13^5
c) 2^6+3^6
d) 2^7+3^8
e) 4^5+9^5

a seemingly easy question, but is there a quick way to get the answer? thanks!

C

2(2^5) + 3(3^5)
2^6 + 3^6

2^5 = X
3^5 = Y

X+X+Y+Y+Y

2X + 3 Y

2(2^5) + 3 (3^5)

(2^1)(2^5) = 2^6 (see rules of exponents)
(3^1)(3^5) = 3^6

Answer
C) 2^6+3^6

i can't improve on the answers given here, but i can point out that this transformation - 2(2^n) = 2^(n+1), and so on (true for all integers, not just "2', but i didn't want to throw too many variables in the same statement) - is EXTREMELY common on the exam. you should get to know it; it will be your best friend on these types of problems.

in general, you should get VERY good at exponent rules - especially at applying them to INTEGERS. many students are extremely good at applying exponent rules to variables and algebraic expressions, but founder when it's time to apply the same rules to whole numbers. because of the prevalence of number properties problems on the test, it's much more likely that you'll have to apply exponent rules to numbers than to variables, so get ready for that.