Question

Find the value of $ {( - 7)^{ - 3}} \times {49^2} \times {11^{ - 4}} \times {121^2} $

Answer 1

Hint: The power is used to express mathematical equations in the short form; it is an expression that represents the repeated multiplication of the same factor. For example - $ 2 \times 2 \times 2 $ can be expressed as $ {2^3} $ . Here, the number two is called the base and the exponent represents the number of times the base is used as the facto

Complete step-by-step answer:
 First convert the given equation in the terms of the prime numbers.
   $ {( - 7)^{ - 3}} \times {49^2} \times {11^{ - 4}} \times {121^2}\,{\text{ }}.....{\text{(1)}} $
   $
\Rightarrow 49 = {7^2}{\text{ }}......{\text{(2)}} \\
\Rightarrow 121 = {11^2}{\text{ }}......{\text{(3)}} \\
    $
Also, by the negative exponent rule- the negative exponents in the numerator when moved to the denominator become positive and vice-versa. Such as $ {a^{ - n}} = \dfrac{1}{{{a^n}}} $
\[
\Rightarrow {( - 7)^{ - 3}} = \dfrac{1}{{{{( - 7)}^3}}} \\
\Rightarrow {( - 7)^{ - 3}} = \dfrac{1}{{( - 7) \times ( - 7) \times ( - 7)}} \\
\Rightarrow {( - 7)^{ - 3}} = - \dfrac{1}{{(7) \times (7) \times (7)}} \\
\Rightarrow {( - 7)^{ - 3}} = - {(7)^{ - 3}}{\text{ }}.....{\text{(4)}} \\
 \]
(By the property- Minus multiplied with multiplied makes it plus and when plus is multiplied with minus gives us minus)
Place values of equation $ (2),\;{\text{(3), (4) in equation (1)}} $
   $
\Rightarrow {( - 7)^{ - 3}} \times {49^2} \times {11^{ - 4}} \times {121^2} \\
   = - {(7)^3} \times {[{(7)^2}]^2} \times {11^{ - 4}} \times {11^4} \\
    $
(Power rule: to raise Power to power you have to multiply the exponents such as - $ {\left( {{2^a}} \right)^b} = {2^{ab}} $ )
   $ = - {(7)^3} \times {(7)^4} \times {11^{ - 4}} \times {11^4} $
Make pair of the terms with the same base
\[ = - \underline {{{(7)}^3} \times {{(7)}^4}} \times \underline {{{11}^{ - 4}} \times {{11}^4}} \]
(Apply product rule to multiply the exponents with the same base, you have to simply add the power such as $ {x^m} \times {x^n} = {x^{m + n}} $ )
   $ = - {(7)^{ - 3 + 4}} \times {11^{ - 4 + 4}} $
Simplify using the basic mathematical operations, remember that addition of terms with one positive and the negative, we have to do subtraction and sign of term with large number)
   $ = - {(7)^1} $
Therefore, the required solution is-
   $ {( - 7)^{ - 3}} \times {49^2} \times {11^{ - 4}} \times {121^2}\, = ( - 7) $

Note: Remember the concept of prime numbers to solve these types of questions. Prime numbers are the natural numbers greater than $ 1 $ and which are not the product of any two smaller natural numbers. $ 1 $ is neither prime nor composite. For Example: $ 2,3,5,7 $ $ ,..... $ $ 2 $ is the prime number as it can have only $ 2 $ factors. The number $ 1 $ (one) and the number itself that is $ 2 $ . Hence, the factors of $ 2 = 2 \times 1 $