Square root of 3^4 or √3^4 A.1 B.2 C.3 D.4

Mathematics

1 answer:

tangare [24]4 years ago

7 0

Answer:

answer is 9.

Step-by-step explanation:

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Can anyone help me!

zhenek [66]

Answer:

$\large \text{$ \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)}$}$

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

Therefore:

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

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

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

Rewrite 256 as 2⁸ :

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

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

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

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

In one complete calculation:

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