This is probably quite simple but I am not sure where to start, any help appreciated!

[(4a^-2)^-2]^-1

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- Nov 23rd 2008, 02:35 PMSqrtof-1isiIndices / powers help
This is probably quite simple but I am not sure where to start, any help appreciated!

[(4a^-2)^-2]^-1 - Nov 23rd 2008, 03:21 PMtdschool
Let's use the formula: x^-n= 1/(x^n). We shall start with outer exponent: 1/[(4a^-2)^-2]^1=1/(4a^-2)^-2. Follow the same formula = 1/{1/(4a^-2)^2. Fraction divide by fraction= fraction multiply by reciprocal; (4a^-2)^2, let's simplify this; 16a^-4, and one more negative exponent: 16/a^4

- Nov 23rd 2008, 03:28 PMSqrtof-1isiThank you
Thanks this one had me stumped with all the extra exponents