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It might help initially if you were to write out the variable "n" times. So if you had A cubed, write A x A x A. If you had A cubed multiplied by A squared, you'd write A x A x A x A x A. Then you'd see straightaway that A was being multiplied by itself 5 times. So you'd have A to the power 5. And of course 5 is the sum of 3 and 2. Does this help at all?

05 March 2012

I hope this makes sense :-)

07 March 2012

Hi tgb101,
Ian's method is perfect. I just want to check that by 'add exponents' you mean in the case of x^2 multiplied by x^3 where Ians method would apply and not x^2 + 3x^2 in which case you would treat it like a+3a.

07 March 2012

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