What is 2[((8-4)^5)/8 ] ?

The solution to 2[((8-4)^5)/8 ] is 256

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simplify 2[((8-4)^5)/8 ]

simplify

2 [\frac{( 8 - 4 ) ^{5}}{8}]

256

Solution steps

2 (\frac{( 8 - 4 ) ^{5}}{8})

Apply rule:

(a) = a

(\frac{( 8 - 4 ) ^{5}}{8}) = \frac{( 8 - 4 ) ^{5}}{8}

= 2 \cdot \frac{( 8 - 4 ) ^{5}}{8}

Cancel

\frac{( 8 - 4 ) ^{5}}{8} : 128

\frac{( 8 - 4 ) ^{5}}{8}

Subtract the numbers:

8 - 4 = 4

= \frac{4 ^{5}}{8}

Factor the number:

8 = 4 \cdot 2

= \frac{4 ^{5}}{4 \cdot 2}

\frac{4 ^{5}}{4} = 4^{4}

\frac{4 ^{5}}{4}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

4^{5} = 4 \cdot 4^{4}

= \frac{4 \cdot 4 ^{4}}{4}

Cancel the common factor:

4

= 4^{4}

= \frac{4 ^{4}}{2}

Factor

4^{4} : 2^{8}

4^{4}

Factor the number:

4 = 2^{2}

= (2^{2})^{4}

Apply exponent rule:

(a^{b})^{c} = a^{b c}, a \geq 0

= 2^{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= 2^{8}

= \frac{2 ^{8}}{2}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

2^{8} = 2 \cdot 2^{7}

= \frac{2 \cdot 2 ^{7}}{2}

Cancel the common factor:

2

= 2^{7}

2^{7} = 128

= 128

= 2 \cdot 128

Multiply the numbers:

2 \cdot 128 = 256

= 256

\frac{32}{40}

(2 \cdot 3)^{2} - 5^{2}

3^{2} - (- 3)^{2}

\frac{42}{12}

256 \cdot 4