Explanation:
In this case, we need to divide two number where
First number, \(n_1=9\times 10^{-5}\)
Second number, \(n_2=3\times 10^{-9}\)
\(\dfrac{n_1}{n_2}=\dfrac{9\times 10^{-5}}{3\times 10^{-9}}\)
We know that, 3 × 3 = 9
So, \(\dfrac{n_1}{n_2}=\dfrac{3\times 10^{-5}}{10^{-9}}\)
Since, \(\dfrac{a^x}{a^y}=a^{x-y}\)
Also, \(a^x.a^y=a^{x+y}\)
Here, \(\dfrac{10^{-5}}{10^{-9}}=a^{-5-(-9)}=10^4\)
\(\dfrac{n_1}{n_2}=3\times 10^{4}\)
So, on dividing two numbers we get \(3\times 10^{4}\). Hence, this is the required solution.
