Question

How do you simplify \[\left( {6 \cdot {{10}^{ - 2}}} \right)\]?

Answer 1

Hint: Here, we will express the numbers \[6\] and \[10\] as a product of prime factors. Then, we will apply different laws of exponents to the factors and simplify it further using the basic mathematical operations to get the required value.

Formula used:
We will use the following formulas:
\[{a^{ - n}} = \dfrac{1}{{{a^n}}}\]
\[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\]
\[{(ab)^m} = {a^m}{b^m}\]

Complete step by step solution:
We are required to simplify \[\left( {6 \cdot {{10}^{ - 2}}} \right)\].
Let us first simplify the given expression by finding the prime factors of \[6\] and \[10\].
We can write \[6\] as \[6 = 2 \times 3\].
Also, we can write \[10\] as \[10 = 2 \times 5\].
So, we can rewrite the given expression as
\[\left( {6 \cdot {{10}^{ - 2}}} \right) = \left( {2 \times 3} \right){\left( {2 \times 5} \right)^{ - 2}}\]
We will use the law \[{a^{ - n}} = \dfrac{1}{{{a^n}}}\]. Therefore, we get
\[ \Rightarrow \left( {6 \cdot {{10}^{ - 2}}} \right) = \dfrac{{\left( {2 \times 3} \right)}}{{{{\left( {2 \times 5} \right)}^2}}}\]
Now using the property\[{(ab)^m} = {a^m}{b^m}\] in the denominator, we get
\[ \Rightarrow \left( {6 \cdot {{10}^{ - 2}}} \right) = \dfrac{{2 \times 3}}{{{2^2} \times {5^2}}}\] …………………..\[\left( 1 \right)\]
Now applying the law of exponent \[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\] to the RHS of above equation, where \[a = 2\], \[m = 1\] and \[n = 2\].
Hence, we get \[\dfrac{{{2^1}}}{{{2^2}}} = {2^{1 - 2}} = {2^{ - 1}}\]. Applying the law of exponent \[{a^{ - n}} = \dfrac{1}{{{a^n}}}\] to \[{2^{ - 1}}\], where \[a = 2\] and \[n = - 1\], we have \[{2^{ - 1}} = \dfrac{1}{2}\]. So, equation \[\left( 1 \right)\] becomes
\[ \Rightarrow \left( {6 \cdot {{10}^{ - 2}}} \right) = \dfrac{3}{{2 \times {5^2}}}\]
Now, in the denominator we have \[2 \times {5^2}\].
We know that \[{5^2} = 5 \times 5 = 25\].
 So, the denominator becomes
\[2 \times {5^2} = 2 \times 25 = 50\].

Hence, the value of \[\left( {6 \cdot {{10}^{ - 2}}} \right)\] is
\[ \Rightarrow \left( {6 \cdot {{10}^{ - 2}}} \right) = \dfrac{3}{{50}}\]

Note:
In the above, we have found out the prime factorization of 6 and 10. Prime factorization is a method of finding factors of a number in terms of prime numbers. When a number is raised to power then the power is known as an exponent and the number is called its base. Here, the exponent is negative, so to make the exponent positive we have to take the reciprocal of the number.
We can also solve the given expression by directly applying the law of exponent \[{a^{ - n}} = \dfrac{1}{{{a^n}}}\] to \[{10^{ - 2}}\], which gives \[\dfrac{1}{{100}}\]. So,
\[6 \cdot {10^{ - 2}} = \dfrac{6}{{100}} = \dfrac{3}{{50}}\]