What is 4^{-6}*4^4*(2^3*2^{-4})^{-1} ?

The solution to 4^{-6}*4^4*(2^3*2^{-4})^{-1} is 1/8

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simplify 4^{-6}4^4(2^3*2^{-4})^{-1}

simplify

4^{- 6} \cdot 4^{4} \cdot (2^{3} \cdot 2^{- 4})^{- 1}

\frac{1}{8}

Decimal Notation

0.125

Solution steps

4^{- 6} \cdot 4^{4} (2^{3} \cdot 2^{- 4})^{- 1}

4^{- 6} \cdot 4^{4} = \frac{1}{16}

= \frac{1}{16} (2^{3} \cdot 2^{- 4})^{- 1}

(2^{3} \cdot 2^{- 4})^{- 1} = 2

= 2 \cdot \frac{1}{16}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{16}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{16}

Cancel the common factor:

2

= \frac{1}{8}

\frac{( a ^{2} b - b ^{3} )}{( 2 ab - 2 b ^{2} )}

(6 x^{2} + 7 x + 2) - (- 9 x^{2} - 10)

\frac{6 x + 2}{x ^{3} - 2 x - 7}

2 \frac{1}{5}

(- 4 x^{2} + 3 x + 3) - (4 x^{2} + 9)

What is 4^{-6}*4^4*(2^3*2^{-4})^{-1} ?
The solution to 4^{-6}*4^4*(2^3*2^{-4})^{-1} is 1/8