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R u l e s o f I n d i c e s S u m m a r y 1. a n x a m = a n + m 2. a n ÷ a m = a n – m 3. 1 Special Results a 0 = 1 a 1 = a 1n = 1 0n = 0 (unless n = 0) a -n = a n n m 4. a m = a m x n = a n 5. 1 n a = n a 6. m m n a = n a m = a n 7. ( a b ) n = a n b n 8. a b n a b n = n © T Madas

Revision on the rules of indices © T Madas

Evaluate the following, giving your final answers as simple as possible: 22 25 = 2 2 + 5 = 27 = 128 03 = x 1 2 81 = 81 = 9 77 ÷ 72 = 7 7 – 2 = 75 1 1 4 3 3 4 3 3 64 27 2-4 = = = = 24 16 3 60 = 1 15 = 1 71 = 7 3 3 2 3 16 = 2 16 = 4 = 64 2 23 = 2 2 x 3 = 26 = 64 1 = 4 2 = 16 1 3 4-2 27 = 3 27 = 3 © T Madas

Evaluate the following, giving your final answers as simple as possible: 23 23 = 2 3 + 3 = 26 = 64 06 = x 1 2 25 = 25 = 5 48 ÷ 43 = 4 8 – 3 = 45 1 1 2 3 3 2 3 3 8 27 5-2 = = = = 52 25 3 40 = 1 1-1 = 1 31 = 3 4 4 3 4 27 = 3 27 = 3 = 81 4 22 = 2 2 x 4 = 28 = 256 1 = 2 3 = 8 1 4 2-3 16 = 4 16 = 2 © T Madas

Test on the Rules of Indices © T Madas

Evaluate the following, giving your final answers as simple as possible: 22 25 = 2 2 + 5 = 27 = 128 03 = x 1 2 81 = 81 = 9 77 ÷ 72 = 7 7 – 2 = 75 1 1 4 3 3 4 3 3 64 27 2-4 = = = = 24 16 3 60 = 1 15 = 1 71 = 7 3 3 2 3 16 = 2 16 = 4 = 64 2 23 = 2 2 x 3 = 26 = 64 1 = 4 2 = 16 1 3 4-2 27 = 3 27 = 3 © T Madas

Evaluate the following, giving your final answers as simple as possible: 23 23 = 2 3 + 3 = 26 = 64 06 = x 1 2 25 = 25 = 5 48 ÷ 43 = 4 8 – 3 = 45 1 1 2 3 3 2 3 3 8 27 5-2 = = = = 52 25 3 40 = 1 1-1 = 1 31 = 3 4 4 3 4 27 = 3 27 = 3 = 81 4 22 = 2 2 x 4 = 28 = 256 1 = 2 3 = 8 1 4 2-3 16 = 4 16 = 2 © T Madas

© T Madas