Question

How do you multiply and simplify $ - \dfrac{{5{m^5}}}{{63n}}.\dfrac{{81mn}}{{{m^7}}} $ ?

Answer 1

Hint: To solve this problem you should basic property of exponent of multiplication and division.
Properties of exponent having same base number:
For multiplication: $ {x^a}.{x^b} = {x^{a + b}} $ .
For division: $ \dfrac{{{x^n}}}{{{x^m}}} = {x^{n - m}} $ .
Properties of exponent having same power different base number:
For multiplication: $ {x^a}.{y^a} = {(x.y)^a} $ .
For division: $ \dfrac{{{x^a}}}{{{y^a}}} = {\left( {\dfrac{x}{y}} \right)^a} $ .

Complete step by step solution:
To solve this problem.
As given in question we have to multiply and after that simplify the answer of multiplication.
So, by multiplication we get,
  $ - \dfrac{{5{m^5}}}{{63n}}.\dfrac{{81mn}}{{{m^7}}} = - \dfrac{{5 \times 81{m^{5 + 1}}n}}{{63{m^7}n}} $
(By using $ {x^a}.{x^b} = {x^{a + b}} $ )
  $ \Rightarrow - \dfrac{{5 \times 81{m^{5 + 1}}n}}{{63{m^7}n}} = - \dfrac{{405{m^6}n}}{{63{m^7}n}} $
To simplify it we divide numerator by denominator. We get,
  $ - \dfrac{{405{m^6}n}}{{63{m^7}n}} = - \dfrac{{45}}{7}{m^{6 - 7}}{n^{1 - 1}} = - \dfrac{{45}}{7}{m^{ - 1}}{n^0} $
( By using $ {x^0} = 1 $ )
We get,
  $ \Rightarrow - \dfrac{{45}}{7}{m^{ - 1}}{n^0} = - \dfrac{{45}}{7}{m^{ - 1}}.1 = - \dfrac{{45}}{{7m}} $
Hence, from above calculation after multiplying and simplifying we get $ - \dfrac{{5{m^5}}}{{63n}}.\dfrac{{81mn}}{{{m^7}}} = - \dfrac{{45}}{{7m}} $ .
So, the correct answer is “ $ - \dfrac{{45}}{{7m}} $ ”.

Note: as we had noticed from above solution. We had first multiplied it then we had simplified it but we can also first simplify each fraction then we can multiply it. As $ {x^0} = 1 $ vice versa of it is also true that means $ 1 = {x^0} $ . So when we simplify any equation we can use it for further simplification. It is used in complex trigonometric and algebraic problems and comes under engineering as well as the physics field.