
How do you simplify \[{\left( { - 5} \right)^{ - 2}}\]?
Answer
453.6k+ views
Hint: Here we need to simplify the above term using the negative Exponent \[{x^{ - m}} = \dfrac{1}{{{x^m}}}\], as the given number is negative number raised to a negative exponent which denotes a reciprocal value in which the rule says that in negative exponent the numerator gets moved to the denominator and become positive exponents and negative exponent in the denominator get moved to the numerator and becomes positive exponents.
Formula Used:
\[{x^{ - m}} = \dfrac{1}{{{x^m}}}\]
Where, \[x\] is the base term and \[m\] is the exponent.
Complete step by step answer:
Here in the above question we can see that the given terms are negative, hence we need to apply a negative exponent rule in which the negative exponents denote a reciprocal value.As per the rule negative exponents in the numerator get moved to the denominator and become positive exponents and negative exponents in the denominator get moved to the numerator and become positive exponents.As we can see here that the power given is a negative exponent function i.e., \[{\left( { - 5} \right)^{ - 2}}\]
In next step we need to move the value to the denominator as shown below:
\[{\left( { - 5} \right)^{ - 2}} = \dfrac{1}{{{{\left( { - 5} \right)}^2}}}\]
Next simplifying the denominator terms, we get
\[{\left( { - 5} \right)^{ - 2}} = \dfrac{1}{{\left( { - 5} \right)\left( { - 5} \right)}}\]
As denominator consists of both negative terms which gives positive product as shown below\[{\left( { - 5} \right)^{ - 2}} = \dfrac{1}{{\left( 5 \right)\left( 5 \right)}}\]
Hence, after simplifying \[{\left( { - 5} \right)^{ - 2}}\] we get \[{\left( { - 5} \right)^{ - 2}} = \dfrac{1}{{25}}\].
Note:To simplify any given value raised with an exponent first we need to find whether the value given is positive or negative based on that apply the exponent rule and start solving. Exponents rules are based on the terms given at the base and power. Hence, exponents are powers or indices and Multiplication of the powers with the same base, adds the powers together when dividing likes bases the powers are subtracted.The law of negative exponents states that, when a number is raised to a negative exponent, we divide 1 by the base raised to a positive exponent.
Formula Used:
\[{x^{ - m}} = \dfrac{1}{{{x^m}}}\]
Where, \[x\] is the base term and \[m\] is the exponent.
Complete step by step answer:
Here in the above question we can see that the given terms are negative, hence we need to apply a negative exponent rule in which the negative exponents denote a reciprocal value.As per the rule negative exponents in the numerator get moved to the denominator and become positive exponents and negative exponents in the denominator get moved to the numerator and become positive exponents.As we can see here that the power given is a negative exponent function i.e., \[{\left( { - 5} \right)^{ - 2}}\]
In next step we need to move the value to the denominator as shown below:
\[{\left( { - 5} \right)^{ - 2}} = \dfrac{1}{{{{\left( { - 5} \right)}^2}}}\]
Next simplifying the denominator terms, we get
\[{\left( { - 5} \right)^{ - 2}} = \dfrac{1}{{\left( { - 5} \right)\left( { - 5} \right)}}\]
As denominator consists of both negative terms which gives positive product as shown below\[{\left( { - 5} \right)^{ - 2}} = \dfrac{1}{{\left( 5 \right)\left( 5 \right)}}\]
Hence, after simplifying \[{\left( { - 5} \right)^{ - 2}}\] we get \[{\left( { - 5} \right)^{ - 2}} = \dfrac{1}{{25}}\].
Note:To simplify any given value raised with an exponent first we need to find whether the value given is positive or negative based on that apply the exponent rule and start solving. Exponents rules are based on the terms given at the base and power. Hence, exponents are powers or indices and Multiplication of the powers with the same base, adds the powers together when dividing likes bases the powers are subtracted.The law of negative exponents states that, when a number is raised to a negative exponent, we divide 1 by the base raised to a positive exponent.
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