How do you write $\dfrac{1}{169}$ as a negative exponent?

Hint: In this question, we are given a fraction and we need to write it in the form of negative exponent. For this we will use the properties of exponents which are ${{a}^{m}}\cdot {{a}^{n}}={{a}^{m+n}}\text{ and }{{a}^{-m}}=\dfrac{1}{{{a}^{m}}}$. We will first write 169 as multiplication of a number with itself and use the first property to write it in exponential form. Then we will use the second property and write the number in a negative exponential form.

Complete step by step answer:
Here we are given the fraction as $\dfrac{1}{169}$. We need to write this number in a negative exponential form i.e. we want to write it in the form as ${{x}^{-m}}$.
As we can see the denominator of the fraction is 169 which has factors as 13 i.e. we have that, 13 times 13 is 169 so we can write the fraction as $\dfrac{1}{13\times 13}$.
Now we know that every integer can be considered as having an exponent as 1. So 13 can be written as ${{13}^{1}}$. Hence the fraction becomes $\dfrac{1}{{{13}^{1}}\times {{13}^{1}}}$.
By the law of exponents we can use ${{a}^{m}}\cdot {{a}^{n}}={{a}^{m+n}}$ on the denominator where the base a is 13. So we get the fraction as $\dfrac{1}{{{13}^{1+1}}}$.
Simplifying the power we get $\dfrac{1}{{{13}^{2}}}$.
Now we need the negative exponent. So let us use another law of exponent according to which ${{a}^{-m}}=\dfrac{1}{{{a}^{m}}}$ so we get $\dfrac{1}{{{13}^{2}}}={{13}^{-2}}$.
As we can see ${{13}^{-2}}$ is of the form ${{x}^{-m}}$ that we required. Hence $\dfrac{1}{169}$ can be written as ${{13}^{-2}}$ as a number having negative exponents.

Note:
Students should keep in mind all the properties of exponents before solving this sum. Note that we can simply take power of 169 as 1 and then use the property ${{a}^{-m}}=\dfrac{1}{{{a}^{m}}}$ to get the fraction in the form ${{169}^{-1}}$ having negative exponent. So this could be a positive answer. Students can get confused between signs in two properties that are ${{a}^{m}}\cdot {{a}^{n}}={{a}^{m+n}}\text{ and }{{a}^{m-n}}=\dfrac{{{a}^{m}}}{{{a}^{n}}}$.