Answer:
[tex]\large \text{$ \sf 2^{-64}$}[/tex]
Step-by-step explanation:
Given:
[tex]\large \text{$ \sf (256)^{-4^{\frac{3}{2}}}$}[/tex]
This reads as: 256 to the power of "-4 to the power of 3/2".
Therefore, we need to deal with the "-4 to the power of 3/2" first.
Rewrite the 4 as 2²:
[tex]\large \text{$ \sf \implies -(2^2)^{\frac{3}{2}}$}[/tex]
[tex]\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:[/tex]
[tex]\large \text{$ \sf \implies -2^{\left(2 \times \frac{3}{2}\right)}$}[/tex]
[tex]\large \text{$ \sf \implies -2^{3}$}[/tex]
Therefore:
[tex]\large \text{$ \sf \implies -2^3=(-2)(-2)(-2) = -8$}[/tex]
Replace "-4 to the power of 3/2" with -8 :
[tex]\large \text{$ \sf \implies 256^{-8}$}[/tex]
Rewrite 256 as 2⁸ :
[tex]\large \text{$ \sf \implies (2^8)^{-8}$}[/tex]
[tex]\textsf{Again, apply exponent rule} \quad (a^b)^c=a^{bc}:[/tex]
[tex]\large \text{$ \sf \implies 2^{(8 \times -8)}$}[/tex]
[tex]\large \text{$ \sf \implies 2^{-64}$}[/tex]
In one complete calculation:
[tex]\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}[/tex]
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