Question

Can anyone help me!​

Answer 1

Answer:

2^-64

Step-by-step explanation:

256^-2*2^3/2

256^-2^3

256^-8

1/256^8

1/2^8*8

1/2^64

2^-64

Answer 2

Answer:

$\large \sf 2^{-64}$

Step-by-step explanation:

Given:

$\large \text{ \sf (256)^{-4^{\frac{3}{2}}} }$

This reads as:  256 to the power of "-4 to the power of 3/2".

Therefore, we need to deal with the "-4 to the power of 3/2" first.

Rewrite the 4 as 2²:

$\large \text{ \sf \implies -(2^2)^{\frac{3}{2}} }$

$\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:$

$\large \text{ \sf \implies -2^{\left(2 \times \frac{3}{2}\right)} }$

$\large \sf \implies -2^{3}$

Therefore:

$\large \sf \implies -2^3=(-2)(-2)(-2) = -8$

Replace "-4 to the power of 3/2" with -8 :

$\large \sf \implies 256^{-8}$

Rewrite 256 as 2⁸ :

$\large \sf \implies (2^8)^{-8}$

$\textsf{Again, apply exponent rule} \quad (a^b)^c=a^{bc}:$

$\large \sf \implies 2^{(8 \times -8)}$

$\large \sf \implies 2^{-64}$

In one complete calculation:

$\large\begin{aligned} \sf (256)^{-4^{\frac{3}{2}}} & = \sf (256)^{-(2^2)^{\frac{3}{2}}}\\& = \sf (256)^{-2^{\left(2 \times \frac{3}{2}\right)}}\\& =\sf (256)^{-2^{3}}\\& = \sf (256)^{-8}\\& \sf = (2^8)^{-8}\\& = \sf 2^{(8 \times -8)}\\& =\sf 2^{-64}\end{aligned}$

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