How do you simplify $(2^{- 1}) (4^{3}) (8^{- 1})$ $(2^-1)(4^3)(8^-1)$ with a base of 2?

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Answer 1 · 2017-06-25T20:35:00

$2^{2}$ $2^2$

Rewrite each exponent with a base of 2. To do this, you need to find an equivalent to the current base.

$2^{- 1}$ $2^-1$ already has a base of two, so we can leave that one alone.
$4^{3}$ $4^3$ is also equal to $2^{2 \cdot 3}$ $2^(2*3)$ , since $4$ $4$ is equal to $2^{2}$ $2^2$
$8^{- 1}$ $8^-1$ is also equal to $2^{3 \cdot - 1}$ $2^(3*-1)$ , since $8$ $8$ is equal to $2^{3}$ $2^3$

Our expression is now:

$(2^{- 1}) (2^{2 \cdot 3}) (2^{3 \cdot - 1})$ $(2^-1)(2^(2*3))(2^(3*-1))$
= $(2^{- 1}) (2^{6}) (2^{- 3})$ $(2^-1)(2^6)(2^-3)$

When multiplying powers with the same base, add the exponents.

Adding the exponents: $- 1 + 6 - 3 = 2$ $-1 + 6 - 3 = 2$

$= 2^{2}$ $= 2^2$
$= 4$ $=4$

If you want your answer as a base of 2, then keep it as $2^{2}$ $2^2$ :)

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