Question

Find the value of: \[\dfrac{2}{{{{\left( {216} \right)}^{\dfrac{{ - 2}}{3}}}}} - \dfrac{4}{{{{\left( {243} \right)}^{ - \dfrac{1}{5}}}}} + \dfrac{1}{{{{\left( {256} \right)}^{\dfrac{{ - 3}}{4}}}}}\].

Answer 1

Hint: We know in exponents a number is multiplied by itself multiple times i.e. \[x\]raised to power\[y\]. Here \[x\] is called base and \[y\]is exponent. It is expressed as \[{x^y}\].Here we will use the law of exponent i.e.\[{a^{ - 1}} = \dfrac{1}{a}\].

Complete step-by-step solution:
Given =\[\dfrac{2}{{{{\left( {216} \right)}^{\dfrac{{ - 2}}{3}}}}} - \dfrac{4}{{{{\left( {243} \right)}^{ - \dfrac{1}{5}}}}} + \dfrac{1}{{{{\left( {256} \right)}^{\dfrac{{ - 3}}{4}}}}}\]
On substituting\[216 = 6 \times 6 \times 6\], \[243 = 3 \times 3 \times 3 \times 3 \times 3\]and \[256 = 4 \times 4 \times 4 \times 4\], we get:
\[ \Rightarrow \dfrac{2}{{{{\left( {6 \times 6 \times 6} \right)}^{\dfrac{{ - 2}}{3}}}}} - \dfrac{4}{{{{\left( {3 \times 3 \times 3 \times 3 \times 3} \right)}^{ - \dfrac{1}{5}}}}} + \dfrac{1}{{{{\left( {4 \times 4 \times 4 \times 4} \right)}^{\dfrac{{ - 3}}{4}}}}}\]
On substituting\[6 \times 6 \times 6 = {6^3}\], \[3 \times 3 \times 3 \times 3 \times 3 = {3^5}\]and \[4 \times 4 \times 4 \times 4 = {4^4}\], we get:
. \[ \Rightarrow \dfrac{2}{{{{\left( 6 \right)}^{3 \times }}^{\dfrac{{ - 2}}{3}}}} - \dfrac{4}{{{{\left( 3 \right)}^{5 \times }}^{\dfrac{{ - 1}}{5}}}} + \dfrac{1}{{{{\left( 4 \right)}^{4 \times }}^{\dfrac{{ - 3}}{4}}}}\]
On cancelling 3 by 3, 5 by 5 and 4 by 4, we get:
\[ \Rightarrow \dfrac{2}{{{{\left( 6 \right)}^{ - 2}}}} - \dfrac{4}{{{{\left( 3 \right)}^{ - 1}}}} + \dfrac{1}{{{{\left( 4 \right)}^{ - 3}}}}\]
On applying \[{a^1} = \dfrac{1}{{{a^{ - 1}}}}\] ,we get:
\[ \Rightarrow 2 \times {\left( 6 \right)^2} - 4{\left( 3 \right)^1} + {\left( 4 \right)^3}\]
\[ \Rightarrow 2 \times 36 - 4 \times 3 + 64\]
\[ \Rightarrow 72 - 12 + 64\]
\[ \Rightarrow 136 - 12\]
\[ \Rightarrow 124\].

Hence our required value is 124.

Note: These are some other laws of exponents that we must remember while solving such questions as they make the calculations easier.
(i) . \[{a^m} \times {a^n} = {a^{m + n}}\].
(ii). \[{a^m} \div {a^n} = {a^{m - n}}\].
(iii). \[{\left( {{a^m}} \right)^n} = {a^{mn}}\].
(iv). \[{a^m} \times {b^m} = {\left( {ab} \right)^m}\].
(v). \[{\left( a \right)^0} = 1\]
(vi). \[{a^m} \div {b^m} = {\left( {\dfrac{a}{b}} \right)^m}\]
\[(vii){a^{ - 1}} = \dfrac{1}{a}\]
We must remember that we can factorize numbers into powers with the help of L.C.M or prime factorization. The method of prime factorization is used to “break down” or express a given number as a product of prime numbers. More so, if a prime number occurs more than once in the factorization, it is usually expressed in exponential form to make it look more compact. Otherwise, we will have a long list of prime numbers being multiplied together.