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Find the value of the following expression:

(38 ⋅ 2−5 ⋅ 90)−2 ⋅ 2 to the power of negative 2 over 3 to the power of 3, whole to the power of 4 ⋅ 328 (5 points)

Write your answer in simplified form. Show all of your steps.

Find the value of the following expression 38 25 902 2 to the power of negative 2 over 3 to the power of 3 whole to the power of 4 328 5 points Write your answe class=

Respuesta :

[tex]\bf \left( 3^8\cdot 2^{-5}\cdot 9^0 \right)^{-2}\cdot \left( \cfrac{2^{-2}}{3^3}\right)^{-4}\cdot 3^{28} \\\\\\ \left( 3^8\cdot \cfrac{1}{2^5}\cdot 1 \right)^{-2}\cdot \left( \cfrac{3^3}{2^{-2}}\right)^{4}\cdot 3^{28}\implies \left( \cfrac{3^8}{2^5}\right)^{-2}\cdot \left( \cfrac{3^3\cdot 2^2}{1}\right)^{4}\cdot 3^{28}[/tex]

[tex]\bf \left( \cfrac{2^5}{3^8}\right)^{2}\cdot \left( \cfrac{3^3\cdot 2^2}{1}\right)^{4}\cdot 3^{28}\implies \left( \cfrac{2^{5\cdot 2}}{3^{8\cdot 2}} \right)\left( 3^{3\cdot 4}\cdot 2^{2\cdot 4} \right)\cdot 3^{28} \\\\\\ \cfrac{2^{10}}{3^{16}}\cdot 3^{12}\cdot 3^{28}\cdot 2^8\implies \cfrac{2^{10}\cdot 2^8\cdot 3^{12}\cdot 3^{28}}{3^{16}}\\\\\\ 2^{10}\cdot 2^8\cdot 3^{12}\cdot 3^{28}\cdot 3^{-16}[/tex]

[tex]\bf 2^{10+8}\cdot 3^{12+28-16}\implies 2^{18}\cdot 3^{24}\implies 262144\cdot 282429536481 \\\\\\ 74037208411275264[/tex]

The value of the given expression is 4

From the question, the given expression is

[tex](3^{8}\ .\ 2^{-5}\ . \ 9^{0})^{-2} \ . \ (\frac{2^{-2} }{3^{3} }) ^{4} \ . \ 3^{28}[/tex]

This becomes

[tex](3^{8} \times 2^{-5} \times 9^{0})^{-2} \times (\frac{2^{-2} }{3^{3} }) ^{4} \times 3^{28}[/tex]

Then, we get

[tex](3^{8} \times 2^{-5} \times 1)^{-2} \times (\frac{2^{-2} }{3^{3} }) ^{4} \times 3^{28}[/tex]

(NOTE: From one of the laws of indices, [tex]x^{0} =1[/tex])

[tex](3^{8} \times 2^{-5} )^{-2} \times (\frac{2^{-2\times 4} }{3^{3 \times 4} }) \times 3^{28}[/tex]

[tex]3^{8\times -2} \times 2^{-5 \times -2} \times \frac{2^{-8} }{3^{12} } \times 3^{28}[/tex]

[tex]3^{-16} \times 2^{10} \times \frac{2^{-8} }{3^{12} } \times 3^{28}[/tex]

This becomes

[tex]3^{-16} \times 3^{28} \times \frac{2^{-8} }{3^{12} } \times 2^{10}[/tex]

[tex]\frac{3^{-16} \times 3^{28}}{3^{12} } \times {2^{-8} \times 2^{10}[/tex]

From the laws indices, we have that

[tex]x^{y} \times x^{z} = x^{y+z}[/tex]

and

[tex]x^{y} \div x^{z} = x^{y-z}[/tex]

Therefore

[tex]\frac{3^{-16} \times 3^{28}}{3^{12} } \times {2^{-8} \times 2^{10} = 3^{-16+28-12} \times 2^{-8+10}[/tex]

[tex]= 3^{0} \times 2^{2}[/tex]

[tex]=1 \times 2^{2}[/tex]

[tex]= 1 \times 4[/tex]

= 4

Hence, the value of the given expression is 4

Learn more here: https://brainly.com/question/17706565

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