
Evaluate: ${\left( { - 3} \right)^{ - 4}}$
Answer
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Hint: We can see this problem is from indices and powers. This number given is having $ - 3$ as base and $ - 4$ as power. As soon as we see the negative sign in power, we should know that we will have to take reciprocal of the number. So, we will first take the reciprocal of the number and remove the negative sign in the power. Then, we will simplify the rest of the expression by calculating the fourth power of the base. The result thus obtained will be the answer to the given question.
Complete step by step answer:
So, the given question requires us to simplify ${\left( { - 3} \right)^{ - 4}}$. This is of the form ${a^{ - x}}$. We know the law of exponents and indices as \[{a^{ - x}} = \dfrac{1}{{{a^x}}}\]. Hence, negative power means that we have to take reciprocal of the number. So, we can simplify the expression as,
$ \Rightarrow {\left( { - 3} \right)^{ - 4}} = \dfrac{1}{{{{\left( { - 3} \right)}^4}}}$
We can factorise $ - 3 = - 1 \times 3$. So, we get,
$ \Rightarrow {\left( { - 3} \right)^{ - 4}} = \dfrac{1}{{{{\left( { - 1 \times 3} \right)}^4}}}$
Now, we know the law of exponents and indices as ${\left( {ab} \right)^n} = {a^n} \times {b^n}$. So, we get,
$ \Rightarrow {\left( { - 3} \right)^{ - 4}} = \dfrac{1}{{{{\left( { - 1} \right)}^4} \times {3^4}}}$
Now, we know that $\left( { - 1} \right)$ raised to any even power is equal to $1$. Hence, we get,
$ \Rightarrow {\left( { - 3} \right)^{ - 4}} = \dfrac{1}{{1 \times {3^4}}}$
Now, we also know that ${3^4} = 81$. So, we get,
$ \therefore {\left( { - 3} \right)^{ - 4}} = \left( {\dfrac{1}{{81}}} \right)$
So, the value of ${\left( { - 3} \right)^{ - 4}}$ can be simplified as $\left( {\dfrac{1}{{81}}} \right)$.
Note: These rules or laws of indices help us to minimize the problems and get the answer in very less time. These powers can be positive and negative but can be moulded according to our convenience while solving the problem. Also note that cube-root, square-root are fractions with 1 as numerator and respective root in denominator.
Complete step by step answer:
So, the given question requires us to simplify ${\left( { - 3} \right)^{ - 4}}$. This is of the form ${a^{ - x}}$. We know the law of exponents and indices as \[{a^{ - x}} = \dfrac{1}{{{a^x}}}\]. Hence, negative power means that we have to take reciprocal of the number. So, we can simplify the expression as,
$ \Rightarrow {\left( { - 3} \right)^{ - 4}} = \dfrac{1}{{{{\left( { - 3} \right)}^4}}}$
We can factorise $ - 3 = - 1 \times 3$. So, we get,
$ \Rightarrow {\left( { - 3} \right)^{ - 4}} = \dfrac{1}{{{{\left( { - 1 \times 3} \right)}^4}}}$
Now, we know the law of exponents and indices as ${\left( {ab} \right)^n} = {a^n} \times {b^n}$. So, we get,
$ \Rightarrow {\left( { - 3} \right)^{ - 4}} = \dfrac{1}{{{{\left( { - 1} \right)}^4} \times {3^4}}}$
Now, we know that $\left( { - 1} \right)$ raised to any even power is equal to $1$. Hence, we get,
$ \Rightarrow {\left( { - 3} \right)^{ - 4}} = \dfrac{1}{{1 \times {3^4}}}$
Now, we also know that ${3^4} = 81$. So, we get,
$ \therefore {\left( { - 3} \right)^{ - 4}} = \left( {\dfrac{1}{{81}}} \right)$
So, the value of ${\left( { - 3} \right)^{ - 4}}$ can be simplified as $\left( {\dfrac{1}{{81}}} \right)$.
Note: These rules or laws of indices help us to minimize the problems and get the answer in very less time. These powers can be positive and negative but can be moulded according to our convenience while solving the problem. Also note that cube-root, square-root are fractions with 1 as numerator and respective root in denominator.
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