Question

By what number should ${{3}^{-2}}$ be divided to get ${{9}^{-2}}?$

Answer 1

Hint: We know the identity given by ${{x}^{-n}}=\dfrac{1}{{{x}^{n}}}.$ We will transform the given numbers as fractions using the given identity. We will change the denominators using the square terms. We know that ${{3}^{2}}=9$ and ${{9}^{2}}=81.$

Complete step by step solution:
We are asked to find the number with which we should divide ${{3}^{-2}}$ to get ${{9}^{-2}}.$
Let us suppose that the number is $a.$ That is, if we divide ${{3}^{-2}}$ with $a,$ we will get ${{9}^{-2}}.$
Let us write the given numbers with negative exponents using only positive exponents with the help of the above identity that helps us to write a number with negative exponent as a number with positive exponent.
We know the identity given by ${{x}^{-n}}=\dfrac{1}{{{x}^{n}}}.$
When we consider ${{3}^{-2}},$ we will get ${{3}^{-2}}=\dfrac{1}{{{3}^{2}}}.$
When we consider ${{9}^{-2}},$ we will get ${{9}^{-2}}=\dfrac{1}{{{9}^{2}}}.$
We know that the square of $3$ is $9$ and the square of $9$ is $81.$
So, we can replace the denominators of the above obtained fractions using the square terms.
We will get $\dfrac{1}{{{3}^{2}}}=\dfrac{1}{9}$ and $\dfrac{1}{{{9}^{2}}}=\dfrac{1}{81}.$
Now, we can say that $a$ divides $\dfrac{1}{9}$ to produce $\dfrac{1}{81}.$
We can write $\dfrac{1}{9}\div a=\dfrac{1}{81}.$
We can write this as $\dfrac{\dfrac{1}{9}}{a}=\dfrac{1}{9a}=\dfrac{1}{81}.$
We will get $\dfrac{1}{9a}=\dfrac{1}{81}.$
Let us take the reciprocal of the whole equation to get $9a=81.$
From this, we will get $a=\dfrac{81}{9}=9.$
Therefore, $a=9.$
So, when we divide $\dfrac{1}{9}$ with $9,$ we will get $\dfrac{1}{81}.$
Hence the required number is $9={{3}^{2}}.$

Note: We can write $a$ as a fraction with denominator $1$ as $a=\dfrac{a}{1}.$ So, we will have $\dfrac{1}{9}\div \dfrac{a}{1}.$ When we divide a fraction with another, we will take the reciprocal of the divisor and multiply it with the dividend. So, we will get $\dfrac{1}{9}\times \dfrac{1}{a}=\dfrac{1}{9a}.$