Question

Express the following in power notation:
\[ - \dfrac{{32}}{{243}}\]

Answer 1

Hint:
Here, we will find the prime factorization of the numerator and the denominator of the fraction. Then we will use formulas of power notation to express the numerator and the denominator in power notation. We will use the formula for the division of numbers with the same power to express the fraction in power notation.

Formulas used:
We will use the following formulas:
If we multiply a number (say \[a\] ) with itself \[n\] times, then this product is represented as \[a\] to the power \[n\] :
\[ \Rightarrow \underbrace {a \times a \times ... \times a}_{n{\rm{ times}}} = {a^n}\]
If two numbers (say \[a\] and \[b\] ) with the same power (say \[n\] ) are divided, they can also be written as their quotient raised to the same power:
\[ \Rightarrow \dfrac{{{a^n}}}{{{b^n}}} = {\left( {\dfrac{a}{b}} \right)^n}\]

Complete step by step solution:
We have to write \[ - \dfrac{{32}}{{243}}\] in power notation. First of all, we will find the prime factorization of numerator 32. We will divide 32 by prime numbers starting from the smallest prime number starting from 2 and write the quotient under it until we reach 1.
Dividing 32 by 2, we get
\[\begin{array}{l}\left. 2 \right|32\\\left. {} \right|16\end{array}\]
The quotient is 16. We can see that 16 is also divisible by 2. So,
\[\begin{array}{l}\left. 2 \right|16\\\left. {} \right|8\end{array}\]
We will repeat the above process.
\[\begin{array}{l}\left. 2 \right|8\\\left. 2 \right|4\\\left. 2 \right|2\\\left. {} \right|1\end{array}\]
We will write all the steps performed above in the form of an integrated table. Therefore, we get
\[\begin{array}{l}\left. 2 \right|32\\\left. 2 \right|16\\\left. 2 \right|8\\\left. 2 \right|4\\\left. 2 \right|2\\\left. {} \right|1\end{array}\]
Expressing 32 as a product of its prime factors, we get
\[32 = 2 \times 2 \times 2 \times 2 \times 2\]
We can see that 32 is equivalent to 2 multiplied by itself 5 times. So, 32 can be written as:
\[32 = {2^5}\] ……………………….\[\left( 1 \right)\]
Now, we will find the prime factorization of denominator 243. We will divide 243 by prime numbers starting from the smallest prime number and write the quotient under it until we reach 1. As 243 has 3 as its unit digit, 243 will not be divisible by 2. The sum of digits of 243 is 9 which is divisible by 3, so 243 is divisible by 3.
We will now divide 243 continuously by 3.
\[\begin{array}{l}\left. 3 \right|243\\\left. 3 \right|81\\\left. 3 \right|27\\\left. 3 \right|9\\\left. 3 \right|3\\\left. {} \right|1\end{array}\]
We will express 243 as a product of its prime factors:
\[243 = 3 \times 3 \times 3 \times 3 \times 3\]
We can see that 243 is equivalent to 3 multiplied by itself 5 times. So, 243 can be written as:
\[243 = {3^5}{\rm{ }}\left( 2 \right)\]
We will substitute equation (1) and equation (2) in the original fraction:
\[ \Rightarrow - \dfrac{{32}}{{243}} = - \dfrac{{{2^5}}}{{{3^5}}}\]
We know that \[\dfrac{{{a^n}}}{{{b^n}}}\] is equal to \[{\left( {\dfrac{a}{b}} \right)^n}\] . We will substitute 2 for \[a\] , 3 for \[b\] and 5 for \[n\] in this formula:
\[ \Rightarrow - \dfrac{{{2^5}}}{{{3^5}}} = - {\left( {\dfrac{2}{3}} \right)^5}\]

$\therefore -\dfrac{32}{243}$ is written as \[ - {\left( {\dfrac{2}{5}} \right)^5}\] in power notation.

Note:
In the notation \[{a^x}\] , \[a\] is known as the base number and \[x\] is known as its power or exponent. Exponents are very useful to understand scientific scales and to write scientific notations.
Here we have prime factorization to solve the equation. Prime factorization is a method of expressing a number in terms of its prime factors. Prime factors contain only prime numbers. Prime numbers are the numbers which has only 2 factors that itself and 1.