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multiplying negative exponents
- Thread starter rhmrpm13
- Start date
If you multiply (-2)×(-3), the result is +6. The second negative power flips the denominator back to the numerator.(a^-2)^-3*(a^5)^2 is my problem. I'm not sure what to do with the negative after the parentesis. I know that it becomes (1/a^2), but I don't know what to do with the -3 outside of that.
There are two ways to attack this.(a^-2)^-3*(a^5)^2 is my problem. I'm not sure what to do with the negative after the parentesis. I know that it becomes (1/a^2), but I don't know what to do with the -3 outside of that.
The long way is to get rid of all negative exponents first and then all possible fractions.
\(\displaystyle \left((a^{(-2)}\right)^{(-3)} = \left(\dfrac{1}{a^2}\right)^{(-3)} = \dfrac{1}{\left(\dfrac{1}{a^2}\right)^3} = \dfrac{1}{1} * \dfrac{\left(a^2\right)^3}{1^3} = \left(a^2\right)^3 = a ^{(2 * 3)} = a^6.\)
The quick way
\(\displaystyle \left(a^{(-2)}\right)^{(-3)} = a^{\{(-2) * (-3)\}} = a^6.\)
There are two ??? ways to attack this.
\(\displaystyle \bigg(a^{-2}\bigg)^{-3} \ = \ \bigg(\dfrac{1}{a^2}\bigg)^{-3} \ = \ \bigg(\dfrac{a^2}{1}\bigg)^3 \ = \ \bigg(a^2\bigg)^3 \ = \ a^{(2\cdot3)} \ = a^6\)