It's basic knowledge, there is no formula. You're expected to know the laws of indices in GCSE, are you not? You will revisit indices at the start of Y12 in Core 1 anyway, but do remember that you can use this trick as it will come in useful throughout AS and A2.

a^b \cdot a^c \equiv a^{b+c}

(edited 7 years ago)

Reply 5

AdeptDz

OP

21

ohhh shame i do know that, my bad just havent done maths for a bit.. thanks anyway

Reply 6

AdeptDz

OP

21

One more thing: can you not * the (k-1) by 2?

Reply 7

AdeptDz

OP

21

why is it just left, but the 2^k and 1 are * by 2

Reply 8

B_9710

18

Original post by AdeptDz

why is it just left, but the 2^k and 1 are * by 2

(k-1)2^k \times 2 =(k-1)2^{k+1}

. This is exactly the same as

ab \times a =a^2 b

, you're saying why isn't b multiplied as well but it is. I think you need to go over the basics of algebra again to clarify some things, because it will be impossible to progress always.

Reply 9

AdeptDz

OP

21

Oh i get it, ye i should.. i feel stupid

OP

Thanks a lot btw

Original post by AdeptDz

Oh i get it, ye i should.. i feel stupid

Unparseable latex formula:

\displaystyle [br]\begin{equation*} a^b \times a^c = \underbrace{a \times a \times \cdots a}_{b \, \text{times}} \times \underbrace{a \times a \times \cdots a}_{c \,\text{times}} = \underbrace{a \times a \times \cdots a}_{(b+c) \, \text{times}} = a^{b+c}\end{equation*}

Reply 12

HapaxOromenon3

3

Original post by Zacken

Unparseable latex formula:

\displaystyle [br]\begin{equation*} a^b \times a^c = \underbrace{a \times a \times \cdots a}_{b \, \text{times}} \times \underbrace{a \times a \times \cdots a}_{c \,\text{times}} = \underbrace{a \times a \times \cdots a}_{(b+c) \, \text{times}} = a^{b+c}\end{equation*}

That's only valid when b and c are positive integers.