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07 Feb 2023, 10:30
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C
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65% (01:59) correct
35%
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If x is a positive integer, then \((-2^x)^{-2} + (2^{-x})^2 =\)
A. \(\frac{1}{2^{−2x}}\)
B. \(\frac{1}{2^{−2x+1}}\)
C. \(2^{(1 -2x)}\)
D. \(2^{-2x}\)
E. \(2^{(-2x -1)}\)
A. \(\frac{1}{2^{−2x}}\)
B. \(\frac{1}{2^{−2x+1}}\)
C. \(2^{(1 -2x)}\)
D. \(2^{-2x}\)
E. \(2^{(-2x -1)}\)
17 Feb 2023, 08:29
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\((−2^x)\)^(−2) + (2^(−x))^2 = ?
=>\(\frac{ 1}{{(-2)^x}^2 }+ [\frac{1}{{2^x}}]^2\)
=>\( \frac{1}{{2^(2x)}} + \frac{1}{{2^(2x)}}\)
=> \(\frac{2}{{2^(2x)}}\)
=> 2^(1−2x)
Answer C
=>\(\frac{ 1}{{(-2)^x}^2 }+ [\frac{1}{{2^x}}]^2\)
=>\( \frac{1}{{2^(2x)}} + \frac{1}{{2^(2x)}}\)
=> \(\frac{2}{{2^(2x)}}\)
=> 2^(1−2x)
Answer C
01 Aug 2024, 04:49
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↧↧↧ Detailed Video Solution to the Problem ↧↧↧
Given that \((-2^x)^{-2} + (2^{-x})^2\) and x is a positive integer
\((-2^x)^{-2} + (2^{-x})^2\) = \(\frac{1}{(-2^x)^{2}} + 2^{-x*2}\)
= \(\frac{1}{2^{x*2}} + 2^{-2x}\)
= \(\frac{1}{2^{2x}} + 2^{-2x}\)
= \(2^{-2x} + 2^{-2x}\)
= \(2^{-2x}*(1 + 1)\)
= \(2^{-2x}*2^1\)
= \(2^{-2x + 1}\)
= \(2^{1 - 2x}\)
So, Answer will be C
Hope it helps!
Watch the following video to MASTER Exponents
Given that \((-2^x)^{-2} + (2^{-x})^2\) and x is a positive integer
\((-2^x)^{-2} + (2^{-x})^2\) = \(\frac{1}{(-2^x)^{2}} + 2^{-x*2}\)
= \(\frac{1}{2^{x*2}} + 2^{-2x}\)
= \(\frac{1}{2^{2x}} + 2^{-2x}\)
= \(2^{-2x} + 2^{-2x}\)
= \(2^{-2x}*(1 + 1)\)
= \(2^{-2x}*2^1\)
= \(2^{-2x + 1}\)
= \(2^{1 - 2x}\)
So, Answer will be C
Hope it helps!
Watch the following video to MASTER Exponents
