Put $> o r <$ sign in the boxes:
${(- 1)}^{27} {▭ (- 1)}^{14}$

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Solution

Comparison of exponents
Given exponential form: ${(- 1)}^{27} a n d {(- 1)}^{14}$
We know the law of exponential is: ${(- 1)}^{n} = 1 i f n i s e v e n$
$= - 1 i f n i s o d d$
We will solve both exponential forms one by one:
${(- 1)}^{27} = - 1 {A s 27 i s o d d}$
${(- 1)}^{14} = 1 {A s 14 i s e v e n}$
Now we can see that $- 1 < 1$
Hence ${(- 1)}^{27} < {(- 1)}^{14}$

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