Is it (b^2)^(-6) all over b^(-4)?
When 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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