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October 31, 2014

October 31, 2014

Posted by **kate** on Monday, November 19, 2007 at 7:29pm.

(b^2)^-6/b^-4

- algebra -
**Michael**, Monday, November 19, 2007 at 7:33pmIs it (b^2)^(-6) all over b^(-4)?

- algebra -
**kate**, Monday, November 19, 2007 at 7:39pmyes

- algebra -
**Michael**, Monday, November 19, 2007 at 7:43pmWhen we have a power raised to another power, we multiply the powers.

(b^2)^(-6) = b^(-12)

Our problem is now b^(-12) over b^(-4).

Since the variable is the same, the variable with the largest power "wins."

If we have b^5 over b^2, that simplifies to b^3. We subtract the losing power (2) from the winning power (5).

In our case, the winning power is -4 because it is larger than -12. We subtract the loser from the winner: -4 - (-12) = -4 + 12 = -8.

We now have b^(-8). To rewrite it with positive exponents only, we simply negate the power and flip it. 1 / b^8 is the final answer.

We can check this. Let b = 3.

(3^2)^(-6) / 3^(-4)

9^(-6) / 3^(-4)

Use your calculator.

1.52 x 10^(-4)

1 / 3^8 is the same thing, so our answer is correct.

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