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If  2x2×8282x=18 \frac{2^{x^2}\times 8^2}{8^{2x}}=\frac{1}{8}\,82x2x2×82​=81​​, then the value of  2(x2−2x+1)2^{(x^2-2x+1)}2(x2−2x+1)​ 
Question

If  2x2×8282x=18 \frac{2^{x^2}\times 8^2}{8^{2x}}=\frac{1}{8}\,​, then the value of  2(x22x+1)2^{(x^2-2x+1)}​ is equal to:

A.

8

B.

2

C.

16

D.

4

Correct option is C

Given:
  2x2×8282x=18 \frac{2^{x^2}\times 8^2}{8^{2x}}=\frac{1}{8}\,​​,
Formula:
am×an=am+n am/an=amna^m × a^n = a^{m+n}\\\ \\a^m / a^n = a^{m−n}​​
Solution:
2x2×82÷82x=18 2x2×(23)2÷(23)2x=23 2x2+66x=23 x26x+6=3x26x+9=0(x3)2=0x=3 2x22x+1=2322(3)+1 =296+1=24162^{x^2}\times 8^2 \div 8^{2x}=\frac{1}{8}\\\ \\2^{x^2}\times (2^3)^2 \div (2^3)^{2x}=2^{-3}\\\ \\2^{x^2+6-6x}=2^{-3}\\\ \\x^2-6x+6=-3\\x^2-6x+9=0\\(x-3)^2=0\\x=3\\\ \\2^{x^2-2x+1}=2^{3^2-2(3)+1}\\\ \\=2^{9-6+1}\\\\=2^4\\\boxed{16}​​
Hence, the correct answer is (c) 16.

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